# Group Theory: General

## Permutations

### Fall 2021 #1#algebra/qual/work

Let $$G$$ be a group. An automorphism $$\phi: G \rightarrow G$$ is called inner if the automorphism is given by conjugation by a fixed group element $$g$$, i.e., \begin{align*} \phi=\phi_{g}: h \mapsto g h g^{-1} . \end{align*}

• Prove that the set of inner automorphisms forms a normal subgroup of the group of all automorphisms of $$G$$.

• Give an example of a finite group with an automorphism which is not inner.

• Denote by $$S_{n}$$ the group of permutations of the set $$\{1, \ldots, n\}$$. Suppose that $$g \in S_{n}$$ sends $$i$$ to $$g_{i}$$ for $$i=1, \ldots, n .$$ Let $$(a, b)$$ denote as usual the cycle notation for the transposition which permutes $$a$$ and $$b$$. For $$i \in\{1, \ldots, n-1\}$$, compute $$\phi_{g}((i, i+1))$$.

• Suppose that an automorphism $$\phi \in \operatorname{Aut}\left(S_{n}\right)$$ preserves cycle type, i.e., that for every element $$s$$ of $$S_{n}, s$$ and $$\phi(s)$$ have the same cycle type. Show that $$\phi$$ is inner.

Hint: Consider the images of generators $$\phi((1,2)), \phi((2,3)), \cdots, \phi((n-1, n))$$.

## Cosets

### Spring 2020 #2#algebra/qual/completed

Let $$H$$ be a normal subgroup of a finite group $$G$$ where the order of $$H$$ and the index of $$H$$ in $$G$$ are relatively prime. Prove that no other subgroup of $$G$$ has the same order as $$H$$.

• Division algorithm: $$(a,b)= d\implies as+bt =1$$ for some $$s, t$$.
• Coset containment trick: $$X\subseteq N \iff xN = N$$ for all $$x$$.

Recognize that it suffices to show $$hN = N$$. Context cue: coprimality hints at division algorithm. Descend to quotient so you can leverage both the order of $$h$$ and the order of cosets simultaneously.

• For ease of notation, replace $$H$$ in the problem with $$N$$ so we remember which one is normal.
• Write $$n\coloneqq{\sharp}N$$ and $$m \coloneqq[G:N] = {\sharp}G/N$$, where the quotient makes sense since $$N$$ is normal.
• Let $$H \leq G$$ with $${\sharp}H = n$$, we’ll show $$H=N$$.
• Since $${\sharp}H = {\sharp}N$$ it suffices to show $$H \subseteq N$$.
• It further suffices to show $$hN = N$$ for all $$h\in H$$.
• Noting $$\gcd(m, n)=1$$, use the division algorithm to write $$1 = ns + mt$$ for some $$s,t\in {\mathbf{Z}}$$.
• The result follows from a computation: \begin{align*} hN &= h^1 N \\ &= h^{ns + mt}N \\ &= h^{ns} N \cdot h^{mt}N \\ &= \qty{h^n N}^s \cdot \qty{h^t N}^m \\ &= (eN)^s \cdot N \\ &= N ,\end{align*}
• We’ve used that $$h\in H \implies o(h) \divides {\sharp}H = n$$ by Lagrange, so $$h^n = e$$.
• We’ve also used that $${\sharp}G/N = m$$, so $$(xH)^m = H$$ for any $$xH\in G/H$$.

### Fall 2014 #6#algebra/qual/completed

Let $$G$$ be a group and $$H, K < G$$ be subgroups of finite index. Show that \begin{align*} [G: H\cap K] \leq [G: H] ~ [G:K] .\end{align*}

• For $$H, K\leq G$$, intersection is again a subgroup of everything: $$H\cap K \leq H, K, G$$ by the one-step subgroup test.
• Counting in towers: $$A\leq B \leq C \implies [C:A] = [C:B][B:A]$$.
• Fundamental theorem of cosets: $$xH = yH \iff xy^{-1}\in H$$.
• Common trick: just list out all of the darn cosets!

Count in towers, show that distinct coset reps stay distinct.

• $$H \cap K \leq H \leq G \implies [G: H \cap K] = [G: H] [H : H \cap K]$$
• So it suffices to show $$[H: H \cap K] \leq [G: K]$$
• Write $$H/H \cap K = \left\{{ h_1 J, \cdots, h_m J }\right\}$$ as distinct cosets where $$J \coloneqq H \cap J$$.
• Then $$h_i J\neq h_j J \iff h_i h_j^{-1}\not\in J = H \cap K$$.
• $$H$$ is a subgroup, so $$h_i h_j^{-1}\in H$$ forces this not to be in $$K$$.
• But then $$h_i K \neq h_j K$$, so these are distinct cosets in $$G/K$$.
• So $${\sharp}G/K \geq m$$.

### Spring 2013 #3#algebra/qual/completed

Let $$P$$ be a finite $$p{\hbox{-}}$$group. Prove that every nontrivial normal subgroup of $$P$$ intersects the center of $$P$$ nontrivially.

\todo[inline]{Clean up, sketchy argument.}


• Let $$N{~\trianglelefteq~}P$$, then for each conjugacy class $$[n_i]$$ in $$N$$, $$H \cap[g_i] = [g_i]$$ or is empty.
• $$G = {\textstyle\coprod}_{i\leq M} [g_i]$$ is a disjoint union of conjugacy classes, and the conjugacy classes of $$H$$ are of the form $$[g_i] \cap H$$.
• Then pull out the center \begin{align*} H = \coprod_{i\leq M} [g_i] \cap H = \qty{ Z(G) \cap H } {\textstyle\coprod}\coprod_{i\leq M'} [g_i] .\end{align*}
• Taking cardinalities, \begin{align*} {\sharp}H = {\sharp}\qty{ Z(G) \cap H} + \sum_{i\leq M'} {\sharp}[g_i] .\end{align*}
• $$p$$ divides $$H$$ since $$H\leq P$$ and $$P$$ is a $$p{\hbox{-}}$$group.
• Each $${\sharp}[g_i] \geq 2$$ since the trivial conjugacy classes appear in the center, forcing $${\sharp}[g_i] \geq p$$.
• $$p$$ divides $${\sharp}[g_i]$$ since $${\sharp}[g_i]$$ must divide $${\sharp}P = p^k$$
• So $$p$$ must divide the remaining term $$Z(G) \cap H$$, which makes it nontrivial.

## Burnside / Class Equation

### Spring 2019 #4#algebra/qual/completed

For a finite group $$G$$, let $$c(G)$$ denote the number of conjugacy classes of $$G$$.

• Prove that if two elements of $$G$$ are chosen uniformly at random,then the probability they commute is precisely \begin{align*} \frac{c(G)}{{\left\lvert {G} \right\rvert}} .\end{align*}

• State the class equation for a finite group.

• Using the class equation (or otherwise) show that the probability in part (a) is at most \begin{align*} \frac 1 2 + \frac 1 {2[G : Z(G)]} .\end{align*}

Here, as usual, $$Z(G)$$ denotes the center of $$G$$.

(DZG) This is a slightly anomalous problem! It’s fun and worth doing, because it uses the major counting formulas. Just note that the techniques used in this problem perhaps don’t show up in other group theory problems.

• Notation: $$X/G$$ is the set of $$G{\hbox{-}}$$orbits
• Notation: $$X^g = \left\{{x\in X{~\mathrel{\Big\vert}~}g\cdot x = x}\right\}$$
• Burnside’s formula: $${\sharp}{X/G} = {1 \over {\sharp}G} \sum {\sharp}{X^g}$$.
• Definition of conjugacy class: $$C(g) = \left\{{ hgh^{-1}{~\mathrel{\Big\vert}~}h\in G }\right\}$$.

Fixed points of the conjugation action are precisely commuting elements. Apply Burnside. Context clue: $$1/[G:Z(G)]$$ is weird, right? Use that $$[G:Z(G)] = {\sharp}G/{\sharp}Z(G)$$, so try to look for $${\sharp}Z(G)/{\sharp}(G)$$ somewhere. Count sizes of centralizers.

• Define a sample space $$\Omega = G \times G$$, so $${\sharp}{\Omega} = ({\sharp}{G})^2$$.

• Identify the event we want to analyze: \begin{align*} A \coloneqq\left\{{(g,h) \in G\times G {~\mathrel{\Big\vert}~}[g,h] = 1}\right\} \subseteq \Omega .\end{align*}

• Note that the slices are centralizers: \begin{align*} A_g \coloneqq\left\{{(g, h) \in \left\{{ g }\right\} \times G {~\mathrel{\Big\vert}~}[g, h] = 1}\right\} = Z(g) \implies A = \coprod_{g\in G} Z(g) .\end{align*}

• Set $$n$$ be the number of conjugacy classes, note we want to show $$P(A) = n / {\left\lvert {G} \right\rvert}$$.

• Let $$G$$ act on itself by conjugation, which partitions $$G$$ into conjugacy classes.

• What are the orbits? \begin{align*} \mathcal{O}_g = \left\{{hgh^{-1}{~\mathrel{\Big\vert}~}h\in G}\right\} ,\end{align*} which is the conjugacy class of $$g$$. In particular, the number of orbits is the number of conjugacy classes.

• What are the fixed points? \begin{align*}X^g = \left\{{h\in G {~\mathrel{\Big\vert}~}hgh^{-1}= g}\right\},\end{align*} which are the elements of $$G$$ that commute with $$g$$, which is isomorphic to $$A_g$$.

• Identifying centralizers with fixed points, \begin{align*} {\sharp}{A} = {\sharp}{\coprod_{g\in G} Z(g) } = \sum_{g\in G} {\sharp}{Z(g)} = \sum_{g\in G}{\sharp}{X^g} .\end{align*}

• Apply Burnside \begin{align*} {\sharp}{X / G} = \frac { 1 } { {\sharp}G } \sum _ { g \in G } {\sharp}X ^ { g } , \end{align*}

• Note $${\sharp}{X/G} = n$$, i.e. the number of conjugacy classes is the number of orbits.

• Rearrange and use definition: \begin{align*} n \cdot {\sharp}{G} = \qty{{\sharp}{X/G} }\cdot {\sharp}{G} = \sum _ { g \in G } {\sharp}X ^ { g } \end{align*}

• Compute probability: \begin{align*} P(A) = {{\sharp}A \over {\sharp}\Omega} = \sum _{ g \in G } \frac{{\sharp}X ^ { g }}{ ( {\sharp}{G} )^2} = \frac{\qty{ {\sharp}{X/G}} \cdot {\sharp}{G}}{ ({\sharp}{G})^2} = \frac{n \cdot {\sharp}{G}}{( {\sharp}{G} )^2} = \frac n {{\sharp}G} .\end{align*}

Statement of the class equation: \begin{align*} {\left\lvert {G} \right\rvert} = Z(G) + \sum_{\substack{\text{One $x$ from each} \\ \text{conjugacy class}}}[G: Z(x)] \end{align*} where $$Z(x) = \left\{{g\in G {~\mathrel{\Big\vert}~}[g, x] = 1}\right\}$$ is the centralizer of $$x$$ in $$G$$.

(DZG): I couldn’t convince myself that a previous proof using the class equation actually works. Instead, I’ll borrow the proof from this note

• Write the event as $$A = \coprod_{g\in G} \left\{{g}\right\} \times Z(g)$$, then \begin{align*} P(A) = {{\sharp}A\over ({\sharp}G)^2} = {1\over ({\sharp}G)^2} \sum_{g\in G} {\sharp}Z(g) .\end{align*}
• Attempt to estimate the sum: pull out central elements $$g\in Z(G)$$.
• Note $$Z(g) = G$$ for central $$g$$, so $${\sharp}Z(g) = {\sharp}G$$
• Note \begin{align*} g\not\in Z(G)\implies {\sharp}Z(g) \leq {1\over 2} {\sharp}G ,\end{align*} since $$Z(g) \leq G$$ is a subgroup, and \begin{align*} [G:Z(g)] \neq 1 \implies [G: Z(g)] \geq 2 .\end{align*}
• Use these facts to calculate: \begin{align*} P(A) &= {1\over ({\sharp}G)^2 } \qty{ \sum_{g\in Z(g)} {\sharp}Z(g) + \sum_{g\not\in Z(g)} {\sharp}Z(g) } \\ &= {1\over ({\sharp}G)^2 } \qty{ \sum_{g\in Z(g)} {\sharp}G + \sum_{g\not\in Z(g)} {\sharp}Z(g) } \\ &= {1\over ({\sharp}G)^2 } \qty{ {\sharp}Z(G) \cdot {\sharp}G + \sum_{g\not\in Z(g)} {\sharp}Z(g) } \\ &\leq {1\over ({\sharp}G)^2 } \qty{ {\sharp}Z(G) \cdot {\sharp}G + \sum_{g\not\in Z(g)} {1\over 2} {\sharp}G } \\ &= {1\over ({\sharp}G)^2 } \qty{ {\sharp}Z(G) \cdot {\sharp}G + \qty{ \sum_{g\not\in Z(g)} {1\over 2} } \cdot {\sharp}G } \\ &= {1\over ({\sharp}G) } \qty{ {\sharp}Z(G) + \sum_{g\not\in Z(g)} {1\over 2} } \\ &= {1\over ({\sharp}G) } \qty{ {\sharp}Z(G) + {1\over 2} \sum_{g\not\in Z(g)} 1 } \\ &= {1\over ({\sharp}G) } \qty{ {\sharp}Z(G) + {1\over 2} {\sharp}(G \setminus Z(G) ) } \\ &= {1\over ({\sharp}G) } \qty{ {\sharp}Z(G) + {1\over 2} {\sharp}G - {1\over 2} {\sharp}Z(G) } \\ &= {1\over ({\sharp}G) } \qty{ {1\over 2} {\sharp}Z(G) + {1\over 2} {\sharp}G } \\ &= {1\over 2} \qty{1 + { {\sharp}Z(G) \over {\sharp}G }} \\ &= {1\over 2} \qty{1 + { 1 \over [G : Z(G)] }} .\end{align*}
\todo[inline]{Redo part c}


## Group Actions / Representations

### Spring 2017 #1#algebra/qual/completed

Let $$G$$ be a finite group and $$\pi: G\to \operatorname{Sym}^*(G)$$ the Cayley representation.

(Recall that this means that for an element $$x\in G$$, $$\pi(x)$$ acts by left translation on $$G$$.)

Prove that $$\pi(x)$$ is an odd permutation $$\iff$$ the order $${\left\lvert {\pi(x)} \right\rvert}$$ of $$\pi(x)$$ is even and $${\left\lvert {G} \right\rvert} / {\left\lvert {\pi(x)} \right\rvert}$$ is odd.

(DZG): This seems like an unusually hard group theory problem. My guess is this year’s qual class spent more time than usual on the proof of Cayley’s theorem.

• $$\operatorname{Sym}(G) \coloneqq\mathop{\mathrm{Aut}}_{\mathsf{Set}}(G, G)$$ is the group of set morphisms from $$G$$ to itself, i.e. permutations of elements of $$G$$.
• More standard terminology: this is related to the left regular representation where $$g\mapsto \phi_g$$ where $$\phi_g(x) = gx$$, regarded instead as a permutation representation.
• This action is transitive!
• Cayley’s theorem: every $$G$$ is isomorphic to a subgroup of a permutation group. In particular, take $$\left\{{ \phi_g {~\mathrel{\Big\vert}~}G\in G }\right\}$$ with function composition as a subgroup of $$\mathop{\mathrm{Aut}}_{\mathsf{Set}}(G)$$.

(DZG): Warning!! I haven’t checked this solution very carefully, and this is kind of a delicate parity argument. Most of the key ideas are borrowed from here.

• Write $$k \coloneqq o(\pi_g)$$, then since $$\pi$$ is injective, $$k = o(g)$$ in $$G$$.
• Since $$\pi_g$$ as a cycle is obtained from the action of $$g$$, we can pick an element $$x_0$$ in $$G$$, take the orbit under the action, and obtain a cycle of length $$k$$ since the order of $$g$$ is $$k$$. Then continue by taking any $$x_1$$ not in the first orbit and taking its orbit. Continuing this way exhausts all group elements and yields a decomposition into disjoint cycles: \begin{align*} \pi_g = (x_0, gx_0, g^2 x_0, \cdots, g^{k-1} x_0) (x_1, gx_1, g^2 x_1, \cdots, g^{k-1} x_1) \cdots (x_m, gx_m, g^2 x_m, \cdots, g^{k-1} x_m) .\end{align*}
• So there are $$m$$ orbits all of length exactly $$k$$. Proceed by casework.
• If $$k$$ is even:
• This yields $$m$$ odd cycles, and thus $$\pi$$ has zero (an even number) of even cycles.
• Thus $$\pi \in \ker \operatorname{sgn}$$ and is an even permutation.
• If $$k$$ is odd
• This yields $$m$$ even cycles, thus an even number of even cycles iff $$m$$ is even
• The claim is that the number of orbit representatives $$m$$ is equal to $$[G:H] = {\sharp}G/H$$ for $$H = \left\langle{ g }\right\rangle$$.
• Proof: define a map \begin{align*} \left\{{ \text{Orbit representatives } x_i }\right\} &\to G/H \\ x &\mapsto xH .\end{align*}
• This is injective and surjective because \begin{align*} xH = yH &\iff xy^{-1}\in H = \left\langle{ g }\right\rangle \\ &\iff xy^{-1}= g^\ell \\ &\iff x=g^\ell y \\ &\iff y\in {\mathcal{O}}_x ,\end{align*} so $$y$$ and $$x$$ are in the same orbit and have the same orbit representative.
• We now have \begin{align*} \pi_g \text{ is an even permutation } \iff \begin{cases} k \text{ is odd and } m \text{ is even} & \\ \text{ or } & \\ k \text{ is even} & . \end{cases} \end{align*}
• Everything was an iff, so flip the evens to odds: \begin{align*} \pi_g \text{ is an odd permutation } \iff \begin{cases} k \text{ is even and } m \text{ is odd} & \\ \text{ or } & \\ k \text{ is odd} & . \end{cases} .\end{align*}
• Then just recall that $$k\coloneqq o(\pi_g)$$ and \begin{align*} m= [G: \left\langle{ g }\right\rangle] = {\sharp}G / {\sharp}\left\langle{ g }\right\rangle= {\sharp}G / o(g) = {\sharp}G/ o(\pi_g) .\end{align*}

### Fall 2015 #1#algebra/qual/completed

Let $$G$$ be a group containing a subgroup $$H$$ not equal to $$G$$ of finite index. Prove that $$G$$ has a normal subgroup which is contained in every conjugate of $$H$$ which is of finite index.

(DZG) A remark: it’s not the conjugates that should be finite index here, but rather the normal subgroup.

• Let $$H\leq G$$ and define $$n\coloneqq[G:H]$$.
• Write $$G/H = \left\{{ x_1 H, \cdots, x_n H }\right\}$$ for the finitely many cosets.
• Let $$G$$ act on $$G/H$$ by left translation, so $$g\cdot xH \coloneqq gxH$$.. Call the action $$\psi: G\to \operatorname{Sym}(G/H)$$.
• Then $${\operatorname{Stab}}(xH) = xHx^{-1}$$ is a subgroup conjugate to $$H$$, and $$K\coloneqq\ker \psi = \bigcap_{i=1}^n xHx^{-1}$$ is the intersection of all conjugates of $$H$$.
• Kernels are normal, so $$K{~\trianglelefteq~}G$$, and $$K\subseteq xHx^{-1}$$ for all $$x$$, meaning $$K$$ is contained in every conjugate of $$H$$.
• The index $$[G:K]$$ is finite since $$G/K \cong \operatorname{im}\psi$$ by the first isomorphism theorem, and $${\sharp}\operatorname{im}\psi \leq {\sharp}\operatorname{Sym}(G/H) = {\sharp}S_n = n! < \infty$$.

## Conjugacy Classes

### Spring 2021 #2#algebra/qual/completed

Let $$H {~\trianglelefteq~}G$$ be a normal subgroup of a finite group $$G$$, where the order of $$H$$ is the smallest prime $$p$$ dividing $${\left\lvert {G} \right\rvert}$$. Prove that $$H$$ is contained in the center of $$G$$.

Solution due to Swaroop Hegde, typed up + modifications added by DZG.

• $$x\in Z(G)$$ iff $${\sharp}C_x = 1$$, i.e. the size of its conjugacy class is one.
• Normal subgroups are disjoint unions of (some) conjugacy classes in $$G$$.
• In fact, this is a characterization of normal subgroups (i.e. $$H$$ is normal iff $$H$$ is a union of conjugacy classes in $$G$$).
• Why: if $$H{~\trianglelefteq~}G$$ then $$ghg^{-1}\in H$$ for all $$g$$, so $$C_h \subseteq H$$ and $$\bigcup_h C_h = H$$. Conversely, if $$H = \bigcup_{h\in H} C_h$$, then $$ghg^{-1}\in C_h \subseteq H$$ and thus $$gHg^{-1}= H$$.
• Orbit stabilizer theorem: $${\sharp}C_g = {\sharp}G/ {\sharp}K_g$$ where $$C_g$$ is the centralizer and $$K_g$$ is the conjugacy class of $$g$$.
• In particular, $${\sharp}C_g$$ divides $${\sharp}G$$.

Show an element $$x$$ is central by showing $${\sharp}C_x = 1$$.

• Let $$p \coloneqq{\sharp}H$$.

• Let $$\left\{{ C_i }\right\}_{i\leq n}$$ be the conjugacy classes in $$G$$, then $$G = {\textstyle\coprod}_{i\leq n} C_i$$

• By the first fact, there is a sub-collection $$\left\{{ C_{i_j}}\right\}_{j\leq k }$$ such that \begin{align*} H = {\textstyle\coprod}_{j\leq k} C_{i_j} .\end{align*}

• The identity is always in a single conjugacy class, so $$C_e = \left\{{ e }\right\}$$.

• Since $$e\in H$$, without loss of generality, label $$C_{i_1} = \left\{{ e }\right\}$$.

• So \begin{align*} H = \coprod_{j\leq k} C_{i_j} = C_{i_1}{\textstyle \coprod} \displaystyle\coprod_{\substack{ j\leq k \\ j\neq 1} } C_{i_j} .\end{align*}

• Take cardinality in the above equation \begin{align*} p = 1 + \sum_{\substack{ j\leq k \\ j\neq 1 }} {\sharp}C_{i_j} .\end{align*}

• So $${\sharp}C_{i_j} \leq p-1$$ for all $$j\neq 1$$.

• Every $${\sharp}C_{i_j}$$ divides $${\sharp}G$$, but $$p$$ was the minimal prime dividing $${\sharp}G$$, forcing $${\sharp}C_{i_j} = 1$$ for all $$j \neq 1$$.

• This rules out $${\sharp}C_{i_j}$$ being a prime less than $$p$$, but also rules out composites: if a prime $$q\divides {\sharp}C_{i_j}$$, then $$q<p$$ and $$q\divides {\sharp}G$$, a contradiction.
• By fact 3, each $$x\in C_{i_j}$$ satisfies $$x\in Z(G)$$.

• $$\cup C_{i_j} = H$$, so $$H \subseteq Z(G)$$.

### Spring 2015 #1#algebra/qual/completed

For a prime $$p$$, let $$G$$ be a finite $$p{\hbox{-}}$$group and let $$N$$ be a normal subgroup of $$G$$ of order $$p$$. Prove that $$N$$ is contained in the center of $$G$$.

• Definition of conjugacy class: $$[x] = \left\{{gxg^{-1}{~\mathrel{\Big\vert}~}g\in G}\right\}$$.
• A conjugacy class $$[x]$$ is trivial iff $$[x] = \left\{{ x }\right\}$$ iff $$x\in Z(G)$$.
• Sizes of conjugacy classes divide the order of the group they live in.
• This is orbit-stabilizer: $$G\curvearrowright G$$ by $$g\cdot x \coloneqq gxg^{-1}$$, so $${\mathcal{O}}(x) = [x]$$. Then $${\sharp}{\mathcal{O}}(x) = {\sharp}G / {\sharp}{\operatorname{Stab}}(x)$$, so $${\sharp}{\mathcal{O}}(x)$$ divides $${\sharp}G$$.

• Use that $$N{~\trianglelefteq~}G \iff N = {\textstyle\coprod}' [n_i]$$ is a disjoint union of (full) conjugacy classes.
• Take cardinalities: \begin{align*} p = {\sharp}N = \sum_{i=1}^m {\sharp}[n_i] = 1 + \sum_{i=2}^m [n_i] .\end{align*}
• The size of each conjugacy class divides the size of $$H$$ by orbit-stabilizer, so $${\sharp}[n_i] \divides p$$ for each $$i$$.
• But the entire second term must sum to $$p-1$$ for this equality to hold, which forces $${\sharp}[n_i] = 1$$ (and incidentally $$m=p-1$$)
• Then $$[n_i] = \left\{{ n_i }\right\} \iff n_i \in Z(G)$$, and this holds for all $$i$$, so $$N \subseteq Z(G)$$.

## Unsorted / Counting Arguments

### Fall 2021 #2#algebra/qual/completed

Give generators and relations for the non-commutative group $$G$$ of order 63 containing an element of order $$9 .$$

• Idea: take a semidirect product involving $$C_9$$ and $$C_7$$. We’ll need some facts: $$\mathop{\mathrm{Hom}}(C_m, C_n) \cong C_d$$ where $$d = \gcd(m, n)$$, and $$\mathop{\mathrm{Aut}}(C_m)\cong C_m^{\times}$$ which has order $$\phi(m)$$ (since one needs to send generators to generators), which can be explicitly calculated based on the prime factorization of $$m$$.

• Some calculations we’ll need:

• $$\mathop{\mathrm{Aut}}(C_9) \cong C_9^{\times}\cong C_{\phi(9)} \cong C_6$$, using that $$\phi(p^k) = p^{k-1}(p-1)$$.
• $$\mathop{\mathrm{Aut}}(C_7) \cong C_7^{\times}\cong C_{\phi(7)}\cong C_6$$ using that $$\phi(p) = p-1$$.
• To get a nonabelian group, we need a nontrivial semidirect product, so look at $$\mathop{\mathrm{Hom}}(G, \mathop{\mathrm{Aut}}(H))$$ in the two possible combinations.

• $$\mathop{\mathrm{Hom}}(C_7, \mathop{\mathrm{Aut}}(C_9)) \cong \mathop{\mathrm{Hom}}(C_7, C_6) \cong C_1 \coloneqq\left\{{e}\right\}$$ using that $$\mathop{\mathrm{Hom}}(C_m, C_n) \cong C_{d}$$ for $$d = \gcd(m, n)$$. So there are no nontrivial homs here, so only the direct product is possible.
• $$\mathop{\mathrm{Hom}}(C_9, \mathop{\mathrm{Aut}}(C_7)) \cong \mathop{\mathrm{Hom}}(C_9, C_6) \cong C_3$$, so use this!
• Note that we don’t have to consider possibilities for $$C_3\times C_3$$, since including this as a factor would yield no elements of order 9.
• So take $$G\coloneqq C_7 \rtimes_\psi C_9$$ for some $$\psi: C_9 \to \mathop{\mathrm{Aut}}(C_7)$$, and we can take the presentation \begin{align*} G = \left\langle{x, y{~\mathrel{\Big\vert}~}x^7, y^9, yxy^{-1}= \psi(x)}\right\rangle .\end{align*}

• It now suffices to find a nontrivial $$\psi: C_7\to C_7$$. Writing it multiplicatively as $$C_7 = \left\langle{x{~\mathrel{\Big\vert}~}x^7}\right\rangle$$, any map that sends $$x$$ to a generator will do. It suffices to choose any $$k$$ coprime to $$7$$, and then take $$\psi(x) \coloneqq x^k$$, which will be another generator.

• So take

\begin{align*} G = \left\langle{x, y{~\mathrel{\Big\vert}~}x^7, y^9, yxy^{-1}= x^2}\right\rangle .\end{align*}

### Fall 2019 Midterm #5#algebra/qual/completed

Let $$G$$ be a nonabelian group of order $$p^3$$ for $$p$$ prime. Show that $$Z(G) = [G, G]$$.

Note: this is a good problem, it tests several common theorems at once. Proof due to Paco Adajar.

Important notations and definitions:

• The center of $$G$$, denoted by $$Z(G)$$, is the subset of elements of $$G$$ which commute with all elements of $$G$$. That is, if $$x \in Z(G)$$, then for all $$g \in G$$, $$gx = xg$$: \begin{align*}Z(G) = \{ x \in G : gx = xg \, \text{for all } g \in G \}.\end{align*}

In fact, $$Z(G)$$ is not just a subset of $$G$$, but a normal subgroup of $$G$$.

• The commutator subgroup of $$G$$, denoted by $$[G, G]$$, is the subgroup of $$G$$ generated by the commutators of $$G$$, i.e., the elements of the form $$ghg^{-1}h^{-1}$$: \begin{align*}[G, G] = \langle ghg^{-1}h^{-1} : g, h \in G \rangle.\end{align*}

The commutator subgroup $$[G,G]$$ is the smallest normal subgroup of $$G$$ whose quotient is abelian. That is, if $$H$$ is a normal subgroup of $$G$$ for which $$G/H$$ is abelian, then $$[G, G] \le H$$.

Moreover, $$G$$ is abelian if and only if $$[G,G]$$ is trivial.

Theorems to remember and know how to prove:

• $$G/Z(G)$$ Theorem: If $$G/Z(G)$$ is cyclic, then $$G$$ is abelian, i.e., $$G/Z(G)$$ is in fact trivial.

• Lagrange’s Theorem: If $$G$$ is a group of finite order and $$H$$ is a subgroup of $$G$$, then the order of $$H$$ divides that of $$G$$.

• One consequence of this is that every group of prime order is cyclic.
• A $$p$$-group (a group of order $$p^n$$ for some prime $$p$$ and some positive integer $$n$$) has nontrivial center.

• A consequence of the theorems above: every group of order $$p^2$$ (where $$p$$ is prime) is abelian.

Since $$Z(G)$$ is a subgroup of $$G$$ and $$|G| = p^3$$, by Lagrange’s theorem, $$|Z(G)| \in \{1, p, p^2, p^3\}$$.

Since we stipulated that $$G$$ is nonabelian, $$|Z(G)| \ne p^3$$. Also, since $$G$$ is a $$p$$-group, it has nontrivial center, so $$|Z(G)| \ne 1$$. Finally, by the $$G/Z(G)$$ theorem, $$|Z(G)| \ne p^2$$: if $$|Z(G)| = p^2$$, then $$|G/Z(G)| = p$$ and so $$G/Z(G)$$ would be cyclic, meaning that $$G$$ is abelian. Hence, $$|Z(G)| = p$$.

Then, since $$|Z(G)| = p$$, we have that $$|G/Z(G)| = p^2$$, and so $$G/Z(G)$$ is abelian. Thus, $$[G, G] \in Z(G)$$. Since $$|Z(G)| = p$$, then $$|[G,G]| \in \{ 1, p\}$$ again by Lagrange’s theorem. If $$|[G,G]| = p$$ then $$[G,G] = Z(G)$$ and we are done. And, indeed, we must have $$|[G,G]| = p$$, because $$G$$ is nonabelian and so $$|[G,G]| \ne 1$$.

### Spring 2012 #2#algebra/qual/completed

Let $$G$$ be a finite group and $$p$$ a prime number such that there is a normal subgroup $$H{~\trianglelefteq~}G$$ with $${\left\lvert {H} \right\rvert} = p^i > 1$$.

• Show that $$H$$ is a subgroup of any Sylow $$p{\hbox{-}}$$subgroup of $$G$$.

• Show that $$G$$ contains a nonzero abelian normal subgroup of order divisible by $$p$$.

• $$p$$ groups have nontrivial centers.
• Definition of maximality and $$p{\hbox{-}}$$groups
• Sylows are conjugate
• $$Z(G) \operatorname{ch}G$$ always.
• Transitivity of characteristic: $$A \operatorname{ch}B$$ and $$B{~\trianglelefteq~}C$$ implies $$A {~\trianglelefteq~}C$$.

Just use maximality for (a). For (b), centers are always abelian, so $$Z(H)$$ is good to consider, just need to ensure it’s normal in $$G$$. Use transitivity of characteristic.

• By definition, $$S\in {\operatorname{Syl}}_p(G) \iff S$$ is a maximal $$p{\hbox{-}}$$subgroup: $$S<G$$ is a $$p{\hbox{-}}$$group, so $${\sharp}S = p^k$$ for some $$k$$, $$S$$ is a proper subgroup, and $$S$$ is maximal in the sense that there are no proper $$p{\hbox{-}}$$subgroups $$S'$$ with $$S \subseteq S' \subseteq G$$.
• Since $${\sharp}H = p^i$$, $$H$$ is a $$p{\hbox{-}}$$subgroup of $$G$$.
• If $$H$$ is maximal, then by definition $$H\in {\operatorname{Syl}}_p(G)$$
• Otherwise, if $$H$$ is not maximal, there exists an $$H' \supseteq H$$ with $$H'\leq G$$ a $$p{\hbox{-}}$$subgroup properly containing $$H$$.
• In this apply the same argument to $$H'$$: this yields a proper superset containment at every stage, and since $$G$$ is finite, there is no infinite ascending chain of proper supersets.
• So this terminates in some maximal $$p{\hbox{-}}$$subgroup $$S$$, i.e. a Sylow $$p{\hbox{-}}$$subgroup.
• So $$H \subseteq S$$ for some $$S\in {\operatorname{Syl}}_p(G)$$.
• All Sylows are conjugate, so for any $$S' \in {\operatorname{Syl}}_p(G)$$ we can write $$S' = gSg^{-1}$$ for some $$g$$.
• Then using that $$H$$ is normal, $$H \subseteq S \implies H = gHg^{-1}\subseteq gSg^{-1}\coloneqq S'$$. So $$H$$ is contained in every Sylow $$p{\hbox{-}}$$subgroup.

• Claim: $$Z(H) \leq H$$ works.
• It is nontrivial since $$H$$ is a $$p{\hbox{-}}$$group and $$p{\hbox{-}}$$groups have nontrivial centers
• It is abelian since $$Z(Z(H)) = Z(H)$$.
• $${\sharp}Z(H) = p^\ell$$ for some $$\ell \leq i$$ by Lagrange
• It thus remains to show that $$Z(H) {~\trianglelefteq~}G$$.
• Use that $$Z(H) \operatorname{ch}H$$ and use transitivity of characteristic to conclude $$Z(H) {~\trianglelefteq~}H$$.
• That $$Z(H) \operatorname{ch}H$$: let $$\psi \in \mathop{\mathrm{Aut}}(H)$$ and $$x=\psi(y)\in \psi(Z(H))$$ so $$y\in Z(H)$$, then for arbitrary $$h\in H$$, \begin{align*} \psi(y)h &= \psi(y) (\psi \circ \psi^{-1})(h) \\ &= \psi( y \cdot \psi^{-1}(h) ) \\ &= \psi( \psi^{-1}(h) \cdot y ) && \text{since } \psi^{-1}(h)\in H, \, y\in Z(H) \\ &= h\psi(y) .\end{align*}
• That $$A \operatorname{ch}B {~\trianglelefteq~}C \implies A{~\trianglelefteq~}C$$:
• $$A\operatorname{ch}B$$ iff $$A$$ is fixed by every $$\psi\in \mathop{\mathrm{Aut}}(B)$$., WTS $$cAc^{-1}= A$$ for all $$c\in C$$.
• Since $$B{~\trianglelefteq~}C$$, the automorphism $$\psi({-}) \coloneqq c({-})c^{-1}$$ descends to an element of $$\mathop{\mathrm{Aut}}(B)$$.
• Then $$\psi(A) = A$$ since $$A\operatorname{ch}B$$, so $$cAc^{-1}= A$$ and $$A{~\trianglelefteq~}C$$.

### Fall 2016 #1#algebra/qual/completed

Let $$G$$ be a finite group and $$s, t\in G$$ be two distinct elements of order 2. Show that subgroup of $$G$$ generated by $$s$$ and $$t$$ is a dihedral group.

Recall that the dihedral groups of order $$2m$$ for $$m\geq 2$$ are of the form \begin{align*} D_{2m} = \left\langle{\sigma, \tau {~\mathrel{\Big\vert}~}\sigma^m = 1 = \tau^2, \tau \sigma = \sigma^{-1}\tau}\right\rangle .\end{align*}

• Suppose $$G = \left\langle{ a, b}\right\rangle$$ with $$a^2 = b^2 = e$$, satisfying some unknown relations.

• Consider $$ab$$. Since $$G$$ is finite, this has finite order, so $$(ab)^n = e$$ for some $$n\geq 2$$.

• Note $$\left\langle{ab, b}\right\rangle \subseteq \left\langle{a, b}\right\rangle$$, since any finite word in $$ab, b$$ is also a finite word in $$a, b$$.

• Since $$(ab)b = ab^2 = a$$, we have $$\left\langle{ab, b}\right\rangle \subseteq \left\langle{a, b}\right\rangle$$, so $$\left\langle{ab, b}\right\rangle = \left\langle{a, b}\right\rangle$$.

• Write $$D_{2n} = F(r, s) / \ker \pi$$ for $$\pi: F(r, s)\to D_{2n}$$ the canonical presentation map.

• Define \begin{align*} \psi: F(r, s) &\to G \\ r &\mapsto ab \\ t &\mapsto b .\end{align*}

• This is clearly surjective since it hits all generators.

• We’ll show that $$ab, a$$ satisfy all of the relations defining $$D_{2n}$$, which factors $$\psi$$ through $$\ker \pi$$, yielding a surjection $$\tilde \psi: D_{2n} \twoheadrightarrow G$$.

• $$(ab)^n = e$$ by construction, $$b^2 = e$$ by assumption, and \begin{align*} b (ab) b^{-1}= babb^{-1}= ba = b^{-1}a^{-1}= (ab)^{-1} ,\end{align*} corresponding to the relation $$srs^{-1}= r^{-1}$$. Here we’ve used that $$o(a) = o(b) = 2$$ implies $$a=a^{-1}, b=b^{-1}$$.
• Surjectivity of $$\tilde \psi$$ yields $$2n = {\sharp}D_{2n} \geq {\sharp}G$$.

• The claim is that $${\sharp}G \geq 2n$$, which forces $${\sharp}G = 2n$$. Then $$\tilde \psi$$ will be a surjective group morphism between groups of the same order, and thus an isomorphism.

• We have $$\left\langle{ ab }\right\rangle\leq G$$, so $$n\divides {\sharp}G$$.
• Since $$b\not\in \left\langle{ ab }\right\rangle$$, this forces $${\sharp}G > n$$, so $${\sharp}G \geq 2n$$.

Remark: see a more direct proof in Theorem 2.1 and Theorem 1.1 here

### Fall 2019 Midterm #1#algebra/qual/completed

Let $$G$$ be a group of order $$p^2q$$ for $$p, q$$ prime. Show that $$G$$ has a nontrivial normal subgroup.

:::

• Write $${\sharp}G = p^2 q$$

• Cases: first assume $$p>q$$, then do $$q<p$$.

• In any case, we have \begin{align*} n_p \divides q &\implies n_p \in \left\{{ 1,q }\right\} \\ \\ n_q \divides p^2 &\implies n_q \in \left\{{ 1, p, p^2}\right\} .\end{align*}

• If $$n_p=1$$ or $$n_q=1$$, we’re done, so suppose otherwise.

• Case 1: $$:p>q$$.

• Using that $$[n_p]_p \equiv 1$$, consider reducing elements in $$\left\{{1, q}\right\} \operatorname{mod}p$$.
• Since $$q<p$$, we just have $$q\operatorname{mod}p = q$$, and as long as $$q\neq 1$$ we have $$q\not\equiv 1\operatorname{mod}p$$. But since $$n_p\neq 1$$ and $$n_p\neq q$$, this is a contradiction. $$\contradiction$$
• Case 2: $$p< q$$:

• Using that $$[n_q]_q \equiv 1$$, consider reducing $$\left\{{1, p, p^2}\right\}\operatorname{mod}q$$.

• Since now $$p<q$$, we have $$p\operatorname{mod}q = p$$ itself, so $$p\operatorname{mod}q \neq 1$$ and we can rule it out.

• The remaining possibility is $$n_q = p^2$$.

• Supposing that $$n_p \neq 1$$, we have $$n_p=q$$, so we can count \begin{align*} \text{Elements from Sylow } q: n_q( {\sharp}S_q - 1) &= p^2(q-1) + 1 ,\end{align*} where we’ve used that distinct Sylow $$q$$s can only intersect at the identity, and although Sylow $$p$$s can intersect trivially, they can also intersect in a subgroup of size $$p$$.

• Suppose all Sylow $$p$$s intersect trivially, we get at least \begin{align*} \text{Elements from Sylow } p: n_p( {\sharp}S_p - 1) &= q(p^2-1) .\end{align*} Then we get a count of how many elements the Sylow $$p$$s and $$q$$s contribute:  \begin{align*} q(p^2-1) + p^2(q-1) + 1 = p^2q - q + p^2q - p^2 + 1 = p^2q + (p^2-1)(q-1)

p^2q = {\sharp}G ,\end{align*} {=html} provided $$(p^2-1)(q-1) \neq 0$$, which is fine for $$p\geq 2$$ since this is at least $$(2^2-1)(3-2) = 3$$ (since $$p<q$$ and $$q=3$$ is the next smallest prime). $$\contradiction$$

• Otherwise, we get two Sylow $$p$$s intersecting nontrivially, which must be in a subgroup of order at least $$p$$ since the intersection is a subgroup of both. In this case, just considering these two subgroups, we get \begin{align*} \text{Elements from Sylow } p: n_p( {\sharp}S_p - 1) &> p^2 + p^2 - p = 2p^2-p -1 .\end{align*} Then a count: \begin{align*} p^2(q-1) + (2p^2-p - 1) + 1 &= p^2 q- p^2 + 2p^2 -p \\ &= p^2 q + p^2 -p \\ &= p^2q + p(p-1) \\ &> p^2q = {\sharp}G ,\end{align*} a contradiction since this inequality is strict provided $$p\geq 2$$. $$\contradiction$$

### Fall 2019 Midterm #4#algebra/qual/work

Let $$p$$ be a prime. Show that $$S_p = \left\langle{\tau, \sigma}\right\rangle$$ where $$\tau$$ is a transposition and $$\sigma$$ is a $$p{\hbox{-}}$$cycle.

#1 #algebra/qual/work #2 #algebra/qual/completed #6 #3 #4 #5