Basics

Summary of useful qual tips:

  • Slightly obvious but good to remember:
    • Subgroups of abelian groups are automatically normal.
    • If \(N\) is normal in \(G\), then \(N\) is normal in any subgroup containing it.
    • If \(N\leq G\) is the unique group of order \({\sharp}N\), then \(N\) is normal (since any conjugate must have the same size).
    • Using the subgroup correspondence: if \(L/H\leq G/H\) then \(L\leq G\) has size \({\sharp}(L/H){\sharp}H\).
  • Sizes and structure:
    • Quotienting by bigger groups yields smaller indices:

Link to Diagram

  • \(x\) is central iff \([x] = \left\{{e}\right\}\).
  • Unions aren’t (generally) subgroups, intersections always are.
  • Coprime order subgroups intersect trivially.
  • Distinct subgroups of order \(p^n, p^m\) can intersect trivially or in subgroups of order \(p^{\ell}\).
  • Conjugacy:
    • Sizes of conjugacy classes divide \({\sharp}G\) (by orbit-stabilizer).
    • Conjugate subgroups have equal cardinality.
    • Normal subgroups absorb conjugacy classes, and are thus unions of conjugacy classes.
    • Reasoning about conjugacy classes: in \(S_n\) they’re precisely determined by cycle type, i.e. a partition of \(n\).
    • Remembering the class equation: for literally any group action \(\phi: G\curvearrowright X\), one has \(X = \mathrm{Fix}(\phi) {\textstyle\coprod}' {\mathrm{Orb}}(x_i)\) as a disjoint union of fixed points and nontrivial orbits, since orbits partition \(X\). Then take your action to be \(G\curvearrowright G\) by \(\phi: g.x\coloneqq gxg^{-1}\) to get \(\mathrm{Fix}(\phi) = Z(G)\) and \({\mathrm{Orb}}(x_i) = \left\{{gx_ig^{-1}}\right\} = [x_i]\) the conjugacy classes. Now apply orbit stabilizer to get \({\mathrm{Orb}}(x) \cong G/{\operatorname{Stab}}(x)\) where \({\operatorname{Stab}}(x) = Z(x) = C_G(x)\) the centralizer.
  • Cosets:
    • Cosets partition a group.
    • Anything dealing with indices \([G:H]\): try just listing the cosets.
    • \(aH = bH \iff ab^{-1}\in H\).
    • Showing subgroup containment: \(K \subseteq H\) iff \(kH = H\) for all \(k\in K\).
  • Sylows:
    • If \(S_p\) is normal, then \(S_p\) is characteristic. This is useful if \(H\leq G\) and \(P\in{\operatorname{Syl}}_p(H)\) is normal in \(H\), then \(P\) is also normal in \(G\).

For any \(p\) dividing the order of \(G\), \(\mathrm{Syl}_p(G)\) denotes the set of Sylow\({\hbox{-}}p\) subgroups of \(G\).

Definitions

An set morphism that is either injective or surjective between sets of the same size is automatically a bijection. It turns out that a group morphism between groups of the same size that is either injective or surjective is automatically a bijection, and the inverse is automatically a group morphism, so bijective group morphisms are isomorphisms.

If \(a, b\in {\mathbf{Z}}\) with \(\gcd(a, b) = d\), then there exist \(s,t\in {\mathbf{Z}}\) such that \begin{align*} as + bt = d .\end{align*}

This \(d\) can be computed using the extended Euclidean algorithm.

Useful context clue! In particular, this works when \(a, b\) are coprime and \(d=1\), since you can write \(x^1 = x^{as + bt} = x^{as}x^{bt}\) to get interesting information about orders of elements. If you see “coprime” in a finite group question, try the division algorithm.

The order of an element \(g\in G\), denoted \(n \coloneqq o(g)\), is the smallest \(n\in {\mathbf{Z}}^{\geq 0}\) such that \(g^n = e\).

Show that the order of any element in a group divides the order of the group.

An expression of the form \(G = \left\langle{S {~\mathrel{\Big\vert}~}R}\right\rangle\) where \(S\) is a set of elements and \(R\) a set of words defining relations means that \(G \coloneqq F[S] / { \operatorname{cl}}_n(R)\) where \(F[S]\) is the free group on the set \(S\) and \({ \operatorname{cl}}_n(R)\) is the normal closure, the smallest normal subgroup of \(F[S]\) containing \(R\).

Finding morphisms between presentations: if \(G\) is presented with generators \(g_i\) with relations \(r_i\) and \(H\) is any group containing elements \(h_i\) also satisfying \(r_i\), there is a group morphism \begin{align*} \phi: G &\to H \\ g_i &\mapsto h_i \quad \forall i .\end{align*} Why this exists: the presentation yields a surjective morphism \(\pi: F(g_i) \to G\) with \(G\cong F(g_i) / \ker \pi\). Define a map \(\psi: F(g_i) \to H\) where \(g_i\mapsto h_i\), then since the \(h_i\) satisfy the relations \(r_i\), \(\ker \pi \subseteq \ker \psi\). So \(\psi\) factors through \(\ker \pi\) yielding a morphism \(F/\ker \pi \to H\).

Subgroups

A subset \(H\subseteq G\) is a subgroup iff

  • Closure: \(HH \subset H\)
  • Identity: \(e\in H\)
  • Inverses: \(g\in H \iff g^{-1}\in H\).

If \(H\subset G\), then \(\left\langle{H}\right\rangle\) is the smallest subgroup containing \(H\): \begin{align*} \left\langle{H}\right\rangle = \cap\left\{{H{~\mathrel{\Big\vert}~}H\subseteq M \leq G}\right\} M = \left\{{ h_1^{\pm 1} \cdots h_n^{\pm 1} {~\mathrel{\Big\vert}~}n\geq 0, h_i \in H}\right\} \end{align*} where adjacent \(h_i\) are distinct.

The commutator subgroup of \(G\) is denoted \([G, G] \leq G\). It is the subgroup generated by all elementary commutators: \begin{align*} [G, G] \coloneqq\left\langle{ aba^{-1}b^{-1}{~\mathrel{\Big\vert}~}a, b\in G }\right\rangle .\end{align*} It is the smallest normal subgroup \(N{~\trianglelefteq~}G\) such that \(G/N\) is abelian, so if \(H\leq G\) and \(G/H\) is abelian, \(H\subseteq [G, G]\).

Note that elements in \([G, G]\) are generally products of elementary commutators, and not elementary themselves, since we’re taking the group generated by them.

If \(H \subseteq G\) and \(a,b\in H \implies ab^{-1}\in H\), then \(H\leq G\). 


    
    
      
        
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          Identity: \(a=b=x\implies xx^{-1}=e\in H\)
        
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          Inverses: \(a=e, b=x \implies x^{-1}\in H\).
        
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          Closure: let \(x, y\in H\), then \(y^{-1}\in H\) by above, so \(xy = x(y^{-1})^{-1}\in H\).
        
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          Show that the intersection of two subgroups is again a subgroup.
    
      
      
        
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          Hint: one-step subgroup test.
        
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  • 
          Show that if \(H\coloneqq C_m, K\coloneqq C_n \leq G\) are cyclic, then \(H \cap K = C_{d}\) where \(d \coloneqq\gcd(m, n)\).
        
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          Show that the intersection of two subgroups with coprime orders is trivial.
        
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          Show that the union of two subgroups \(H, K\) is a subgroup iff \(H \subset K\), and so is generally not a subgroup.
        
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          Show that subgroups with the same prime order are either equal or intersect trivially.
        
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          Important for Sylow theory: show (perhaps by example) that if \(S_1, S_2\) are distinct subgroups of order \(p^k\), then it’s possible for their intersection to be trivial or for them to intersect in a subgroup of order \(p^\ell\) for \(1\leq \ell \leq k-1\).
        
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          Give a counterexample where \(H,K\leq G\) but \(HK\) is not a subgroup of \(G\).
        
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Conjugacy

    

      The conjugacy class of \(h\) is defined as 

\begin{align*}
C(h) \coloneqq\left\{{ ghg^{-1}{~\mathrel{\Big\vert}~}g\in G }\right\} 
.\end{align*}

    

  

    

      \([e] = \left\{{ e }\right\}\) is always in a conjugacy class of size one – this is useful for counting and divisibility arguments. Conjugacy classes are not subgroups in general, since they don’t generally contain \(e\). However, by orbit-stabilizer and the conjugation action, their sizes always divide the order of \(G\).
    

  
    

      Useful qual fact: \([x] = \left\{{ x }\right\} \iff x\in Z(G)\), i.e. having a trivial conjugacy class is the same as being central.
    

  

    

      Two subgroups \(H, K \leq G\) are conjugate iff there exists some \(g\in G\) such that \(gHg^{-1}= K\). Note that all conjugate subgroups have the same cardinality.
    

  

    

      Show that the size of a conjugacy class divides the order of a group.
    

  

    

      Show that if \(H < G\) is a proper subgroup, then \(\bigcup_{g\in G} gHg^{-1}\subset G\) is a proper subset.
    

  
    

      
    

      Hint: consider the intersection and count. Try Orbit-stabilizer?
    

  
    

  

    

      Strategy: bound the cardinality. All conjugates of \(H\) have the same cardinality, say \({\sharp}H = m\). Suppose there are \(n\) distinct conjugates of \(H\). Then they intersect only at the identity, so count their elements: 

\begin{align*}
{\sharp}\bigcup_{g\in G} gHg^{-1}= 1 + n(m-1)
.\end{align*}
 Use that \(n = [G: N_G(H)]\) by Orbit-Stabilizer, and \(N_G(H) \leq G \implies n \leq n' \coloneqq[G:H]\). Now note \(n'm = {\sharp}H[G:H] = {\sharp}G\) by Lagrange: 

\begin{align*}
{\sharp}\bigcup_{g\in G} gHg^{-1}
&= 1 + n(m-1) \\
&\leq 1 + n'(m-1) \\
&= 1 + n'm -n' \\
&= 1 + {\sharp}G - n' \\
&= {\sharp}G - (n' - 1) \\
&< {\sharp}G && \iff n' \coloneqq[G:H] > 1
.\end{align*}

    

  

    

      Show that normal groups absorb conjugacy classes: if \(N{~\trianglelefteq~}G\) and \([g_i]\) is a conjugacy class in \(g\), either \([g_i] \subseteq N\) or \([g_i] \cap N = \emptyset\).
    

  

    

      Prove that the size of a conjugacy class of \(g_i\) is the index of its centralizer, \([G: Z(g_i)] \coloneqq[G: C_G(g_i)]\).
    

  
Normal Subgroups

    

      A subgroup \(N\leq G\) is normal iff \(gH = Hg\) for every \(g\in G\), or equivalently \(gHg^{-1}= H\) for all \(g\), so \(H\) has only itself as a conjugate. We denote this by \(N{~\trianglelefteq~}G\). Equivalently, for every inner automorphism \(\psi \in \mathop{\mathrm{Inn}}(G)\), \(\psi(N) = N\).
    

  

    

      \(N{~\trianglelefteq~}G \iff N = {\textstyle\coprod}' [h_i]\) is a disjoint union of conjugacy classes, where the index set for this union is one \(h_i\) from each conjugacy class.
    

  

    

      Note that \(C(h_i) = \left\{{ gh_i g^{-1}{~\mathrel{\Big\vert}~}g\in G }\right\}\), and \(gh_i g^{-1}\in H\) since \(H\) is normal, so \(C(h_i) \subseteq G\) for all \(i\). Conversely, if \(C(h_i) \subseteq H\) for all \(h_i \in H\), then \(gh_ig^{-1}\in H\) for all \(i\) and \(H\) is normal.
    

  

    
    
      
        
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          Show that if \(H, K {~\trianglelefteq~}G\) and \(H\cap K = \emptyset\), then \(hk=kh\) for all \(h\in H,k\in K\).
        
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          Show that if \(H,K{~\trianglelefteq~}G\) are normal subgroups that intersect trivially, then \([H, K] = 1\) (so \(hk = kh\) for all \(k\) and \(h\)).
        
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      Prove that if \(G\) is a \(p{\hbox{-}}\)group, every subgroup \(N{~\trianglelefteq~}G\) intersects the center \(Z(G)\).
    

  
    

      
    

      Hint: use the class equation.
    

  
    

  

    

      Easy solution:
    

  
    
    
      
        
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          Use that \({\sharp}H \operatorname{mod}p = 1\) since \(H\leq G\) and \(G\) is a \(p{\hbox{-}}\)group.
        
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          Then use that \(H\) is a union of conjugacy classes, and since \(e\in H\) there is at least one class of size 1, so 

\begin{align*}
{\sharp}H = {\sharp}{\textstyle\coprod}' [h_i] = {\sharp}[e] + \sum' {\sharp}[h_i] \\
\implies 0 \equiv {\sharp}H \equiv 1 + \sum' {\sharp}[h_i] \operatorname{mod}p
,\end{align*}
 and since each \({\sharp}[h_i]\) divides \({\sharp}H\), not all can be of size \(p^\ell\) since then the sum would be \(0\operatorname{mod}p\). So at least one other \({\sharp}[h_i] = 1\), making that \(h_i\) central.
        
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      Another solution:
    

  
    
    
      
        
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      Idea: use the class equation to force \(p\) to divide \({\sharp}(H \cap Z(G))\). Applying it to \(H\) yields 

\begin{align*}
H = Z(H) {\textstyle\coprod}_{i=1}^m [h_i]
,\end{align*}
 where the \([h_i]\) are conjugacy classes of size greater than 1.
    
    
  
        
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      Now use that \(Z(H) = Z(G) \cap H\), and since \(p\) divides the LHS the result will follow if \(p\) divides the size of the disjoint union on the RHS.
    
    
  
        
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      This is true because each \({\sharp}[h_i] \neq 1\) and \([h_i]\) divides \({\sharp}H\) which divides \({\sharp}G\) which is a power of \(p\). So \(p\divides {\sharp}[h_i]\) for each \(i\).
    
    
  
        
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Centralizing and Centers

    

      The centralizer of an element is defined as 

\begin{align*}
Z(h) \coloneqq C_G(h) \coloneqq\left\{{ g\in G {~\mathrel{\Big\vert}~}ghg^{-1}= h }\right\} 
,\end{align*}
 the elements of \(G\) the stabilize \(h\) under conjugation.
    

  
    

      The centralizer of a subset \(H\) is defined as 

\begin{align*}
Z(H) \coloneqq C_G(H) \coloneqq\bigcap_{h\in H} C_G(h) \coloneqq\left\{{g\in G {~\mathrel{\Big\vert}~}ghg^{-1}= h ~\forall h\in H}\right\}
,\end{align*}
 the elements of \(G\) that simultaneously stabilize all of \(H\) pointwise under conjugation.
    

  

    

      

\begin{align*}
N_G(H) = \left\{{g\in G {~\mathrel{\Big\vert}~}gHg^{-1}= H}\right\} = \bigcup_{M\in S} M, \quad S \coloneqq\left\{{H{~\mathrel{\Big\vert}~}H {~\trianglelefteq~}M \leq G}\right\} 
\end{align*}
 Contrast to the centralizer: these don’t have to fix \(H\) pointwise, but instead can permute elements of \(H\).
    

  

    

      \(C_G(H) {~\trianglelefteq~}N_G(H)\) for any \(H\). The main difference between these is that \(C_G(S)\) has to centralize \(H\) pointwise, where \(N_G(H)\) allows the weaker condition of centralizing \(H\) as a set (potentially permuting elements within \(H\)).
    

  
    

      This is maybe easier to remember for Lie algebras: there 

\begin{align*}
C_{{\mathfrak{g}}}({\mathfrak{h}}) = \left\{{x\in {\mathfrak{g}}{~\mathrel{\Big\vert}~}[xh] = 0 \,\forall h\in {\mathfrak{h}}}\right\}
N_{{\mathfrak{g}}}({\mathfrak{h}}) = \left\{{x\in {\mathfrak{g}}{~\mathrel{\Big\vert}~}[xh] \in {\mathfrak{h}}\,\forall h\in {\mathfrak{h}}}\right\}
.\end{align*}
 So \([x, {-}]_{{\mathfrak{h}}} = 0\) for central \(x\) and \(\operatorname{im}[x, {-}]_{{\mathfrak{h}}} \subseteq {\mathfrak{h}}\) for normal elements.
    

  

    

      The center of \(G\) is defined as 

\begin{align*}
Z(G) = \left\{{ g\in G {~\mathrel{\Big\vert}~}[g, h] = e \, \forall h\in H}\right\}
= \left\{{ g\in G {~\mathrel{\Big\vert}~}Z(g) = G }\right\} 
,\end{align*}
 the subgroup of central elements: those \(g\in G\) that commute with every element of \(G\).
    

  

    
    
      
        
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          Show that if \(G/Z(G)\) is cyclic then \(G\) is abelian.
        
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          Show that \(G/N\) is abelian iff \([G, G] \leq N\).
        
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          Show that normal subgroups are not necessarily contained in \(Z(G)\).
    
      
      
        
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          Hint: consider the order 3 subgroup of \(S_3\).
        
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      The \(G/Z(G)\) theorem:
    

  
    
    
      
        
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          Write \(H\coloneqq Z(G)\) and \(G/H = \left\langle{xH}\right\rangle\) as a cyclic quotient.
        
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          Fix \(a, b\in G\), then \(aH = x^n H\) and \(bH = x^m H\).
        
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          So \(ax^{-n} = h_1, bx^{-m} = h_2\) where the \(h_i\) are now central.
        
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          Now write \(ab = (x^n h_1)(x^m h_2) = ba\) by commuting everything.
        
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Cosets

    

      

\begin{align*}
K\leq H \leq G \implies [G: K] = [G:H] [H: K]
.\end{align*}

    

  

    

      If \(H\leq K \leq G\), then 

\begin{align*}
{\sharp}G = [G:1] \geq [G:H] \geq [G:K] \geq [G:G] = 1
.\end{align*}
 In particular, If \(H, K\leq G\) are just arbitrary, since \(H \cap K \leq H, K\) we have \([H: H \cap K] \geq [G:H] \text{ and } [G:K]\).
    

  

    

      Write \(G/H \cap K \coloneqq G/J = \left\{{ h_1J, \cdots, h_m J }\right\}\) as distinct cosets where \(m\coloneqq[G:H]\) and the \(h_i\) are all in \(H\). Then \(i\neq j\implies h_i h_j^{-1}\not \in H \cap K\), but \(h_i h_j^{-1}\in H\) since subgroups are closed under products and inverses, which forces \(h_i h_j^{-1}\not\in K\). So \(h_i K \neq h_j K\), meaning there are at least \(m\) cosets in \(G/K\), so \([G:K] \geq m\).
    

  

    

      Any two cosets \(xH, yH\) are either equal or disjoint.
    

  

    
    
      
        
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      \(x\in xH\), since \(e\in H\) because \(H\) is a subgroup and we can take \(h=e\) to get \(x = xe \coloneqq xh \in xH\).
    
    
  
        
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      The reverse containment is clear, so \(G = \bigcup_{x\in G} xH\) is a union of its cosets.
    
    
  
        
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      Suppose toward a contradiction that \(\ell \in xH \cap yH\) we’ll show \(xH = yH\).
    
    
  
        
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      Write \(\ell =xh_1 =yh_2\) for some \(h_i\), then 

\begin{align*}
xh_1 = yh_2 &\implies x = yh_2 h_1^{-1}\\
xh_3\in xH &\implies xh_3 = (yh_2h_1^{-1}) h_3 \in yH
,\end{align*}
 so \(xH \subseteq yH\).
    
    
  
        
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      A symmetric argument shows \(y_H \subseteq xH\).
    
    
    1
    
  
    
    
  
        
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\begin{align*}
aH = bH \iff a^{-1}b \in H \iff b^{-1}a\in H
.\end{align*}

    

  

    

      
    
    
    1
    
   

\begin{align*}
aH = bH \iff a\in bH \iff a=bh \text{ for some } h \iff b^{-1}a = h \iff ba^{-1}\in H
.\end{align*}

    

  

    

      The index \([G: H]\) of a subgroup \(H\leq G\) is the number of left (or right) cosets \(gH\).
    

  

    

      If you can reduce a problem to showing \(X \subseteq H\), it suffices to show \(xH = H\) for all \(x\in X\).
    

  

    

      Cosets form an equivalence relation and thus partition a group. Nice trick: write \(G/H = \left\{{ g_1 H, g_2 H,\cdots, g_n H }\right\}\), then \(G = {\textstyle\coprod}_{i\leq n} g_i H\).
    

  

    

      If \(H{~\trianglelefteq~}G\) and \(G\) is finite then 

\begin{align*}
[G: H] = {\left\lvert {G/H} \right\rvert} = {{\left\lvert {G} \right\rvert} \over {\left\lvert {H} \right\rvert}}
.\end{align*}

    

  

    

      Show that if \(G\) is finite then \({\left\lvert {G} \right\rvert}/{\left\lvert {H} \right\rvert} = [G: H]\).
    

  
Special Groups

    

      A dihedral group of order \(2n\) is given by 

\begin{align*}
D_n = \left\langle{r, s {~\mathrel{\Big\vert}~}r^n, s^2, rsr^{-1}= s^{-1}}\right\rangle = \left\langle{r, s {~\mathrel{\Big\vert}~}r^n, s^2, (rs)^2 }\right\rangle
\end{align*}
 The \(r\) is a cycle of length \(n\), and \(s\) is a reflection.
    

  
    

      Examples of explicit cycle presentations:
    

  
    
    
      
        
  • 
          \(D_4 = \left\langle{(1,2,3,4), (1,3)}\right\rangle\) which is a \(2\pi/4\) rotation and a reflection through the diagonal line \(y=-x\) in a square.
        
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  • 
          \(D_5 = \left\langle{(1,2,3,4,5), (1,5)(2,4)}\right\rangle\) which is a \(2\pi/5\) rotation and a reflection about the line through vertex \(3\) in a pentagon.
        
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      The Quaternion group of order 8 is given by 

\begin{align*}
Q &= \left\langle{x,y,z {~\mathrel{\Big\vert}~}x^2 = y^2 = z^2 = xyz = -1}\right\rangle \\
&= \left\langle{x, y {~\mathrel{\Big\vert}~}x^4 = y^4, x^2 = y^2, yxy^{-1}= x^{-1}}\right\rangle
\end{align*}
 Mnemonic: multiply clockwise to preserve sign, counter-clockwise to negate sign. Everything squares to \(-1\), and the triple product is \(-1\):
    

  
    

      
 

    

  
    

      
    

      Link to Diagram
    

  
    

  

    

      A subgroup \(H\leq S_n\) is transitive iff its action on \(\left\{{1, 2, \cdots, n}\right\}\) is transitive, i.e. for each pair \((i, j)\) there is some element \(\sigma\in H\) such that \(\sigma(i) = j\). Note that \(\sigma\) may not fix other elements, and can have other effects!
    

  

    

      If \({\left\lvert {G} \right\rvert} = p^k\), then \(G\) is a p-group.
    

  
Cyclic Groups

    

      \(G\) is cyclic of order \(n \coloneqq{\sharp}G\) iff \(G\) has a unique subgroup of order \(d\) for each \(d\) dividing \(n\).
    

  

    

      \(\impliedby\): Use that \(\sum_{d\divides n} \phi(d) = n\), and that there are at most \(\phi(d)\) elements of order \(d\), forcing equality.
    

  
    

      \(\implies\): If \(G = \left\langle{ a }\right\rangle\) with \(a^n=e\), then for each \(d\divides n\) take \(H_d \coloneqq\left\langle{ a^{n\over d} }\right\rangle\) for existence.
    

  

    
    
      
        
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          Show that any cyclic group is abelian.
        
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          Show that every subgroup of a cyclic group is cyclic.
        
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          Show that 

\begin{align*}\phi(n) = n \prod_{p\mathrel{\Big|}n}\qty{1 - {1\over p}}.\end{align*}

        
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          Compute \(\mathop{\mathrm{Aut}}({\mathbf{Z}}/n{\mathbf{Z}})\) for \(n\) composite.
        
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          Compute \(\mathop{\mathrm{Aut}}(\qty{{\mathbf{Z}}/p{\mathbf{Z}}}^n)\).
        
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Symmetric Groups

    

      The transposition presentation: 

\begin{align*}
S_n \coloneqq\left\langle{ \sigma_1, \cdots, \sigma_{n-1} {~\mathrel{\Big\vert}~}\sigma_i^2, [\sigma_i, \sigma_j]\, (j\neq i+1), \sigma_i \sigma_{i+1} \sigma_i = \sigma_{i+1} \sigma_i \sigma_{i+1} }\right\rangle 
.\end{align*}

    

  

    

      Writing a cycle as a product of transpositions, the map defined by 

\begin{align*}
\operatorname{sgn}: S_n &\to ({\mathbf{Z}}/2, +) \\
\prod_{i\leq 2k} (a_i b_i) &\mapsto 0 \\
\prod_{i\leq 2k+1} (a_i b_i) &\mapsto 1
.\end{align*}

    

  

    
    
      
        
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          The kernel is the alternating group:
    
      
      
        
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          Even cycles
        
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          For a single cycle: has odd length
        
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          Have an even number of even length cycles.
        
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          Can be written as an even number of transpositions.
        
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          Examples: \((1,2,3)\) or \((1,2)(3,4)\) in \(S_4\).
        
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          Non-examples: \((1,2)\) or \((1,2,3,4)\) in \(S_4\), since they have an odd number of even length cycles.
        
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          The fiber over 1 is everything else:
    
      
      
        
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          Odd cycles
        
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          For a single cycle: has even length
        
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          Have an odd number of even length cycles.
        
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          Can be written as an odd number of transpositions
        
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      Mnemonic: the cycle parity of a \(k{\hbox{-}}\)cycle is the usual integer parity of \(k-1\).
    

  
    

  

    

      The alternating group is the subgroup of even permutations, i.e. 

\begin{align*}
A_n \coloneqq\left\{{\sigma \in S_n {~\mathrel{\Big\vert}~}\operatorname{sgn}(\sigma) = 0}\right\}
.\end{align*}
 These are the permutations with an even number of even length cycles.
    

  

    

      For \(n\geq 3\), \(A_n\) is generated by 3-cycles.
    

  

    

      Every 3-cycle \((abc)\) is even, and thus in \(A_n\). Given an arbitrary even permutation \((t_1\ldots t_{2k})\), it decomposes into a product of an odd number of transpositions \((t_{2j-1} t_{2j})\). So it suffices to write every such transposition as a 3-cycle. There are only 3 cases the occur:
    

  
    
    
      
        
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          \((ab)(ab) = ()\)
        
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          \((ab)(ac) = (abc)\)
        
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          \((ab)(cd) = (abc)(adc)\).
        
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\begin{align*}
A_3 =
\left\{{ \operatorname{id}, (1,2,3), (1,3,2) }\right\} 
,\end{align*}
 which has cycle types \((1,1,1)\) and \((3)\).
    

  
    

      

\begin{align*}
A_4 =
& \{\operatorname{id}, \\
& (1,3)(2,4),
(1,2)(3,4),
(1,4)(2,3), \\
& (1,2,3),
(1,3,2), \\
& (1,2,4),
(1,4,2), \\
& (1,3,4),
(1,4,3), \\
& (2,3,4),
(2,4,3) \}
,\end{align*}
 which has cycle types \((1,1,1,1), (2,2), (3, 1)\).
    

  
    

      \(A_5\) is too big to write down, but has cycle types
    

  
    
    
      
        
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          \((1,1,1,1,1)\)
        
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          \((2,2,1)\)
        
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          \((3,1,1)\)
        
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          \((5)\)
        
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          \(\sigma \circ (a_1 \cdots a_k)\circ \sigma^{-1} = (\sigma(a_1), \cdots \sigma(a_k))\)
        
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          Conjugacy classes are determined by cycle type
        
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          The order of a cycle is its length.
        
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          The order of an element is the least common multiple of the sizes of its disjoint cycles.
        
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          Disjoint cycles commute.
        
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          \(A_{n\geq 5}\) is simple.
        
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Exercises

    
    
      
        
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      Show that if \(G\) is a finite group acting transitively on a set \(X\) with at least two elements, then there exists \(g \in G\) which fixes no point of \(X\).
    
    
  
        
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      Let \(p\) be prime. For each abelian group \(K\) of order \(p^2\), how many subgroups \(H\leq {\mathbf{Z}}^{\times 3}\) are there with \({\mathbf{Z}}^3/H \cong K\)?
    
    
  
        
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      Let \({\sharp}G = pq\), with \(p,q\) distinct primes. Show that \(G\) has a nontrivial proper normal subgroup, and if \(p\not\equiv 1\operatorname{mod}q\) and \(q\not \equiv 1 \operatorname{mod}p\) then \(G\) is abelian.
    
    
  
        
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      Let \(G\) be a finite group and let \(p\) be the smallest prime dividing \({\sharp}G\), and assume \(G\) has a normal subgroup of order \(p\). Show that \(H \subset Z(G)\).
    
    
  
        
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      Let \(G\) be finite and \(P\) a Sylow 2-subgroup. Assume \(P\) is cyclic and generated by \(x\). Show that the sign of the permutation of \(G\) corresponding to \(x\mapsto gx\) is 1, and deduce that \(G\) has a nontrivial quotient of order 2.
    
    
  
        
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Footnotes

      
        

          

            1.
          

          

            See full argument: D&F p.80.