Automorphism Groups
Homs among various cyclic groups \(C_m\) and any of their automorphism groups \(\mathop{\mathrm{Aut}}(C_m)\) are completely classified, so for example \(\mathop{\mathrm{Hom}}(C_m, C_n), \mathop{\mathrm{Hom}}(C_m, \mathop{\mathrm{Aut}}(C_n)), \mathop{\mathrm{Hom}}(\mathop{\mathrm{Aut}}(C_m), C_n)\), etc. There’s a good reference here:
https://www.whitman.edu/documents/Academics/Mathematics/SeniorProject_BrianSloan.pdf
Let \(\varphi\) be the totient function, and note that a cyclic group \(C_n\) has precisely \(\phi(n)\) choices of generators. One can compute \begin{align*} \phi(p) &= p-1 \\ \phi(p^k) &= p^{k-1}(p - 1) \\ \phi(p^kq^\ell) &= \phi(p^k)\phi(q^\ell) \quad\text{when } \gcd(q, p) = 1 .\end{align*}
- Automorphisms of cyclic groups are completely known: \begin{align*} \mathop{\mathrm{Aut}}(C_n) \cong C_n^{\times} ,\end{align*} which has size \(\phi(n)\) but is not generally isomorphic to \(C_{\phi(n)}\)
Warning: \(C_n^{\times}\) is not always cyclic!! For example, \(C_8^{\times}\cong C_2^2 \neq C_{4}\). In fact, \(C_n^{\times}\) cyclic iff \(n=2,4,p^k, 2p^k\) for \(p\) an odd prime.
-
For \(p\) an odd prime, \(\mathop{\mathrm{Aut}}(C_p) \cong C_p^{\times}\cong C_{p-1}\) is cyclic.
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For \(p^k\) an odd prime power, \(\mathop{\mathrm{Aut}}(C_{p^k}) \cong C_{\varphi(p^k)}\) is cyclic.
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For \(2^k\) with \(k\geq 1\), \(C_{2^k}^{\times}\cong C_{2}\times C_{2^{k-2}}\).
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If \(G, H\) have coprime order then \(\mathop{\mathrm{Aut}}(G \times H) \cong \mathop{\mathrm{Aut}}(G) \times\mathop{\mathrm{Aut}}(H)\). One can then compute a general order by factoring \(n = \prod_{k=1}^\ell p_k^{n_k}\) to get a decomposition \begin{align*} C_n= C_{\prod_{k=1}^\ell p_k^{n_k}}= \prod_{k=1}^{\ell} C_{p_k^{n_k}} ,\end{align*} and thus \begin{align*} \mathop{\mathrm{Aut}}(C_n) &\cong \mathop{\mathrm{Aut}}\qty{\prod_{k=1}^{\ell} C_{p_k^{n_k}} }\\ &\cong \prod_{k=1}^\ell \mathop{\mathrm{Aut}}\qty{C_{p_k^{n_k}}} \\ &\cong \prod_{k=1}^\ell C_{p_k^{n_k}}^{\times}\\ &\cong C_{2^{\ell}}^{\times}\times\prod_{\substack{k=1 \\ p_k\neq 2} }^\ell C_{p_k^{n_k}}^{\times}\\ &\cong \qty{C_2 \times C_{2^{\ell-2}} } \times\prod_{\substack{k=1 \\ p_k\neq 2} }^\ell C_{m_k} && m_k \coloneqq\varphi(p_k^{n_k}) \\ &\cong \qty{C_2 \times C_{2^{\ell-2}} } \times\prod_{\substack{k=1 \\ p_k\neq 2} }^\ell C_{m_k} && m_k \coloneqq p_k^{n_k-1}(p_k-1) .\end{align*}
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\(\mathop{\mathrm{Aut}}(C_p^n) \cong \operatorname{GL}_n({ \mathbf{F} }_p)\) which has size \begin{align*} {\sharp}\operatorname{GL}_n({ \mathbf{F} }_p) = \prod_{k=0}^{n-1}(p^n-p^k) = (p^n-1)(p^n-p)(p^n-p^2)\cdots(p^n-p^{n-1}) .\end{align*}
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\(\mathop{\mathrm{Aut}}(C_m^n)\) for \(m\) not prime: no clue! For \(n=2\), this seems to be a wreath product \(\mathop{\mathrm{Aut}}(C_m) \wr C_2\).
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Counting homs: \({\sharp}\mathop{\mathrm{Hom}}_{\mathsf{Grp}}(C_n, C_m) = \gcd(n ,m)\).
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If \(\sigma \in \mathop{\mathrm{Aut}}(H)\) and \(\tau \in \mathop{\mathrm{Aut}}(N)\), then \(N \rtimes_\psi H \cong N \rtimes_{\tau \circ \psi \circ \sigma} H\).
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So if \(\operatorname{GL}_n\) shows up in a semidirect product, it suffices to consider similarity classes of matrices (i.e. just use canonical forms).
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\(\mathop{\mathrm{Inn}}(G) \cong G/Z(G)\).
Some examples of writing automorphism groups as products of cyclic groups: \begin{align*} \mathop{\mathrm{Aut}}(C_{2^2\cdot 3}) &\cong \mathop{\mathrm{Aut}}(C_{2^2})\times\mathop{\mathrm{Aut}}(C_3) \cong \qty{C_2 \times C_{2^{2-2}}} \times C_{\phi(3)} = \qty{C_2}\times C_2 \\ \mathop{\mathrm{Aut}}(C_{2^3\cdot 3}) &\cong \mathop{\mathrm{Aut}}(C_{2^3})\times\mathop{\mathrm{Aut}}(C_3) \cong \qty{C_2 \times C_{2^{3-2}}}\times C_{\phi(3)} = \qty{C_2 \times C_2}\times C_2 \\ \mathop{\mathrm{Aut}}(C_{2^4\cdot 3}) &\cong \mathop{\mathrm{Aut}}(C_{2^4})\times\mathop{\mathrm{Aut}}(C_3) \cong \qty{C_2 \times C_{2^{4-2}}}\times C_{\phi(3)} = \qty{C_2 \times C_{2^2}}\times C_2\\ \mathop{\mathrm{Aut}}(C_{2^2\cdot 7}) &\cong \mathop{\mathrm{Aut}}(C_{2^2})\times\mathop{\mathrm{Aut}}(C_7) \cong \qty{C_2 \times C_{2^{2-2}}}\times C_{\phi(7)}\cong C_2 \times C_6 \\ \mathop{\mathrm{Aut}}(C_{2\cdot 3\cdot 5}) &\cong \mathop{\mathrm{Aut}}(C_2) \times\mathop{\mathrm{Aut}}(C_3) \times\mathop{\mathrm{Aut}}(C_5) \cong 1\times C_{\phi(3)}\times C_{\phi(5)} \cong C_2 \times C_4 .\end{align*}
Some concrete examples of \(\mathop{\mathrm{Aut}}(C_m) \cong C_m^{\times}\) for various \(m\):
- \(m=2: 0\)
- \(m=3: C_2\)
- \(m=4: C_2\)
- \(m=5: C_2\)
- \(m=6: C_2\)
- \(m=7: C_6\)
- \(m=8: C_2\times C_2\)
- \(m=9: C_6\)
- \(m=10: C_4\)
- \(m=11: C_{10}\)
- \(m=12: C_{2} \times C_{2}\)
- \(m=13: C_{12}\)
- \(m=14: C_{6}\)
- \(m=15: C_{4} \times C_{2}\)
- \(m=16: C_{4} \times C_{2}\)
- \(m=17: C_{16}\)
- \(m=18: C_{6}\)
- \(m=19: C_{18}\)
- \(m=20: C_{4} \times C_{2}\)
- \(m=21: C_{6} \times C_{2}\)
- \(m=22: C_{10}\)
- \(m=23: C_{22}\)
- \(m=24: C_{2} \times C_{2} \times C_{2}\)
- \(m=25: C_{20}\)
- \(m=26: C_{12}\)
- \(m=27: C_{18}\)
- \(m=28: C_{6} \times C_{2}\)
- \(m=29: C_{28}\)
- \(m=30: C_{4} \times C_{2}\)
- \(m=31: C_{30}\)
- \(m=32: C_{8} \times C_{2}\)
- \(m=33: C_{10} \times C_{2}\)
- \(m=34: C_{16}\)
- \(m=35: C_{12} \times C_{2}\)
- \(m=36: C_{6} \times C_{2}\)
- \(m=37: C_{36}\)
- \(m=38: C_{18}\)
- \(m=39: C_{12} \times C_{2}\)
- \(m=40: C_{4} \times C_{2} \times C_{2}\)
- \(m=41: C_{40}\)
- \(m=42: C_{6} \times C_{2}\)
- \(m=43: C_{42}\)
- \(m=44: C_{10} \times C_{2}\)
- \(m=45: C_{12} \times C_{2}\)
- \(m=46: C_{22}\)
- \(m=47: C_{46}\)
- \(m=48: C_{4} \times C_{2} \times C_{2}\)
- \(m=49: C_{42}\)
- \(m=50: C_{20}\)
Isomorphism Theorems
If \(\phi:G\to H\) is a group morphism then \begin{align*}G/\ker \phi \cong \operatorname{im}\phi.\end{align*}
Note: for this to make sense, we also have
- \(\ker \phi {~\trianglelefteq~}G\)
- \(\operatorname{im}\phi \leq G\)
If \(\phi: G\to H\) is surjective then \(H\cong G/\ker \phi\).
If \(S \leq G\) and \(N {~\trianglelefteq~}G\), then \begin{align*} \frac{SN}{N} \cong \frac{S}{S\cap N} \quad \text{ and }\quad {\left\lvert {SN} \right\rvert} = \frac{{\left\lvert {S} \right\rvert} {\left\lvert {N} \right\rvert}}{{\left\lvert {S\cap N} \right\rvert}} .\end{align*}
For this to make sense, we also have
- \(SN \leq G\),
- \(S\cap N {~\trianglelefteq~}S\),
If we relax the conditions to \(S, N \leq G\) with \(S \in N_G(N)\), then \(S\cap N {~\trianglelefteq~}S\) (but is not normal in \(G\)) and the 2nd Isomorphism Theorem still holds.
Suppose \(N, K \leq G\) with \(N {~\trianglelefteq~}G\) and \(N\subseteq K \subseteq G\).
- If \(K\leq G\) then \(K/N \leq G/N\) is a subgroup
- If \(K{~\trianglelefteq~}G\) then \(K/N {~\trianglelefteq~}G/N\).
- Every subgroup of \(G/N\) is of the form \(K/N\) for some such \(K \leq G\).
- Every normal subgroup of \(G/N\) is of the form \(K/N\) for some such \(K {~\trianglelefteq~}G\).
- If \(K{~\trianglelefteq~}G\), then we can cancel normal subgroups: \begin{align*} \frac{G/N}{K/N} \cong \frac{G}{K} .\end{align*}
Suppose \(N {~\trianglelefteq~}G\), then there exists a correspondence:
\begin{align*} \left\{ H < G {~\mathrel{\Big\vert}~}N \subseteq H \right\} \rightleftharpoons \left\{ H {~\mathrel{\Big\vert}~}H < \frac G N \right\} \\ \left\{{\substack{ \text{Subgroups of $G$} \\ \text{containing $N$} }}\right\} \rightleftharpoons \left\{{\substack{ \text{Subgroups of the } \\ \text{quotient $G/N$} }}\right\} .\end{align*}
In words, subgroups of \(G\) containing \(N\) correspond to subgroups of the quotient group \(G/N\). This is given by the map \(H \mapsto H/N\).
\(N {~\trianglelefteq~}G\) and \(N \subseteq H < G \implies N {~\trianglelefteq~}H\).
Products
If \(H,K \leq G\) and \(H \leq N_G(K)\) (or \(K {~\trianglelefteq~}G\)) then \(HK \leq G\) is a subgroup.
\begin{align*} \gcd(p, q) = 1 \implies {\mathbf{Z}}/p{\mathbf{Z}}\times{\mathbf{Z}}/q{\mathbf{Z}}\cong {\mathbf{Z}}/pq{\mathbf{Z}} .\end{align*}
We have \(G \cong H \times K\) when
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\(H, K {~\trianglelefteq~}G\)
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\(G = HK\).
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\(H\cap K = \left\{{e}\right\} \subset G\)
Note: can relax to \([h,k] = 1\) for all \(h, k\).
Prove the “recognizing direct products” theorem. Can the conditions be relaxed?
Things are particularly nice when the orders of \(H\) and \(k\) are coprime. For 3, \(x\in H \cap K\) implies that the order of \(x\) divides \(\gcd({\sharp}H, {\sharp}K) = 1\), so \(H \cap K = \left\{{e}\right\}\). Thus for 2, one only needs that \({\sharp}(HK) = {\sharp}G\).
With these conditions, the following map is an isomorphism: \begin{align*} \Gamma: H\times K &\to G \\ (h, k) &\mapsto hk .\end{align*}
- This is a group morphism by condition (1): \begin{align*} \Gamma(h_1, k_1) \Gamma(h_2, k_2) &\coloneqq(h_1 k_1) (h_2 k_2) = h_1 ({ \color{red} k_1 h_2 } ) k_2 \\ &= h_1 ( { \color{red} h_2 k_1 } ) k_2 \\ &= (h_1 h_2) ( k_1 k_2) \\ &\coloneqq\Gamma( (h_1, k_1)(h_2, k_2) ) .\end{align*}
- This is surjective by condition (2)
- This is injective by condition(3) and checking the kernel: \begin{align*} \ker \Gamma = \left\{{ (h,k) {~\mathrel{\Big\vert}~}hk = 1_G,\, hk = 1_G}\right\} \implies h = k ^{-1} \implies hk \in K \cap H = \left\{{1_G}\right\} .\end{align*}
We have \(G \cong \prod_{i=1}^n H_i\) when
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\(H_i {~\trianglelefteq~}G\) for all \(i\).
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\(G = H_1 \cdots H_n\)
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\(H_k \cap H_1 \cdots \widehat{H_k} \cdots H_n = \emptyset\)
Note on notation: intersect \(H_k\) with the amalgam leaving out \(H_k\).
We have \(G \cong N \rtimes_\psi H\) when
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\(N {~\trianglelefteq~}G\)
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\(G = NH\)
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\(H \curvearrowright N\) by conjugation via a map \begin{align*} \psi: H \to \mathop{\mathrm{Aut}}(N) \\ h \mapsto h({-})h^{-1} .\end{align*}
Relaxed condition: \(H, N {~\trianglelefteq~}G\) for direct product, or just \(H\leq G\) for a semidirect product.
Classification: Finitely Generated Abelian Groups
If \(G\) is a finitely generated abelian group, then there is a decomposition \begin{align*} G \cong {\mathbf{Z}}^r \times \prod_{k=1}^m C_{n_k} \quad \text{ where } n_1 \divides \cdots \divides n_m ,\end{align*} into a free group and a finite number of cyclic groups, where \(r\in {\mathbf{Z}}^{\geq 0}\) is unique and the \(n_i\) are uniquely determined.
If \(G\) is a finitely generated abelian group, then there is a unique list of not necessarily distinct prime powers \(p_k^{e_k}\) such that \begin{align*} G \cong {\mathbf{Z}}^r \times\prod_{k=1}^m C_{p^k}^{e_k} ,\end{align*} where \(r\in {\mathbf{Z}}^{\geq 0}\) is uniquely determined.
Given any presentation of a group as a product of cyclic groups \(G = \prod {\mathbf{Z}}_i/m_i\), with the \(m_i\) not necessarily distinct,
- Factor all of the \(m_i\) into prime powers, keeping the exponents intact.
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Organize into a table whose columns correspond to individual primes \(p_i\).
- Within an individual column for the prime \(p_k\), write all terms of the form \(p_k^{e_k}\) (with exponents intact)
- Arrange the terms from lowest at the top to highest at the bottom. Push everything down so that the bottom-most rows are all filled out.
- For elementary divisors, just list out all of elements of the table individually, running across rows.
- For invariant factors, iterate a process of taking the largest of each prime power (i.e. the bottom row) at each step, deleting that row, and continuing in the same fashion.
Note: this sounds much more complicated than it actually is. Try it!
Suppose \(G\) is given to you as a product of cyclic groups whose sizes factor in the following way \begin{align*} p_1^{e_1}p_1^{e_2}p_1^{e_3} \cdot p_2^{f_1}p_2^{f_2} \cdot p_3^{g_1} p_3^{g_2} p_3^{g_3} \cdot p_4^{h_1} .\end{align*}
Assemble these into a table, grouped by prime factor \(p_i\), being careful not to separate primes from their exponents:
| \(p_1\) | \(p_2\) | \(p_3\) | \(p_4\) |
|---|---|---|---|
| \(p_1^{e_1}\) | \(p_3^{g_1}\) | ||
| \(p_1^{e_2}\) | \(p_2^{f_1}\) | \(p_3^{g_2}\) | |
| \(p_1^{e_3}\) | \(p_2^{f_2}\) | \(p_3^{g_3}\) | \(p_4^{h_1}\) |
For elementary divisors: take columns, which just amounts to listing them again: \begin{align*} &\quad {\mathbf{Z}}/p_1^{e_1} \times{\mathbf{Z}}/p_1^{e_2} \times{\mathbf{Z}}/p_1^{e_3} \\ &\quad \times{\mathbf{Z}}/p_2^{f_1} \times{\mathbf{Z}}/p_2^{f_2} \\ &\quad \times{\mathbf{Z}}/p_3^{g_1} \times{\mathbf{Z}}/p_3^{g_2} \times{\mathbf{Z}}/p_3^{g_3} \\ &\quad \times{\mathbf{Z}}/p_4^{h_1} .\end{align*}
For invariant factors: take rows (grouped by CRT) \begin{align*} & \quad {\mathbf{Z}}/ \qty{p_1^{e_3} p_2^{f_2} p_3^{g_3} p_4^{h_1}} \\ &\quad \times{\mathbf{Z}}/ \qty{p_1^{e_2} p_2^{f_1} p_3^{g_2}} \\ &\quad \times{\mathbf{Z}}/\qty{p_1^{e_1} p_3^{g_1} } .\end{align*}
\begin{align*} G = {{\mathbf{Z}}_{2}\times{\mathbf{Z}}_{2}\times{\mathbf{Z}}_{2}\times{\mathbf{Z}}_{3}\times{\mathbf{Z}}_{3}\times{\mathbf{Z}}_{5^2}} \end{align*}
I’ll use a shortcut for the table: instead of listing columns, I just list the prime powers for a single \(p\) in increasing order in the same cell. Then just always take the largest prime power in each cell at each stage:
| \(p = 2\) | \(p= 3\) | \(p =5\) |
|---|---|---|
| \(2,2,2\) | \(3,3\) | \(5^2\) |
\(\implies n_m = 5^2 \cdot 3 \cdot 2\)
| \(p = 2\) | \(p= 3\) | \(p =5\) |
|---|---|---|
| \(2,2\) | \(3\) | \(\emptyset\) |
\(\implies n_{m-1} = 3 \cdot 2\)
| \(p = 2\) | \(p= 3\) | \(p =5\) |
|---|---|---|
| \(2\) | \(\emptyset\) | \(\emptyset\) |
\(\implies n_{m-2} = 2\)
and thus the invariant factor form is \begin{align*} G\cong {\mathbf{Z}}_2 \times {\mathbf{Z}}_{3\cdot 2} \times {\mathbf{Z}}_{5^2 \cdot 3 \cdot 2} \end{align*}
\begin{align*} G \coloneqq{\mathbf{Z}}_2 \times {\mathbf{Z}}_2 \times {\mathbf{Z}}_{2^3} \times {\mathbf{Z}}_{5^2\cdot 7} \end{align*} Make the table by factoring the order of each cyclic piece, being careful not to combine terms that come from distinct summands (e.g. not combining the two copies of \(2^1\)), and to keep exponents from factorizations intact as a single term (e.g. the \(2^3\)):
| \(2\) | \(5\) | \(7\) |
|---|---|---|
| \(2\) | ||
| \(2\) | ||
| \(2^3\) | \(5^2\) | \(7\) |
Reading across rows from bottom to top (and using CRT to merge everything within a row) yields invariant factors on the LHS below. Reading down columns, left to right (merging nothing) yields elementary divisors on the right-hand side below
\begin{align*} {\mathbf{Z}}_2 \times {\mathbf{Z}}_2 \times {\mathbf{Z}}_{2^3 \cdot 5^2 \cdot 7} \cong {\mathbf{Z}}_2 \times {\mathbf{Z}}_2 \times {\mathbf{Z}}_{2^3} \times {\mathbf{Z}}_{5^2} \times {\mathbf{Z}}_7 .\end{align*}
If \({\sharp}G \coloneqq n = \prod_{k=1}^m p_k^{e_k}\), then there are exactly \(\prod_{k=1}^m P(e_k)\) abelian groups of order \(n\), where \(P\) is the integer partition function.
One can compute \(P(6) = 11\), where all of the partitions are given by \begin{align*} &[6], \\ &[5, 1], \\ &[4, 2], \\ &[4, 1, 1], \\ &[3, 3], \\ &[3, 2, 1], \\ &[3, 1, 1, 1], \\ &[2, 2, 2], \\ &[2, 2, 1, 1], \\ &[2, 1, 1, 1, 1], \\ &[1, 1, 1, 1, 1, 1] .\end{align*}
In practice, it is easy to list all of the partitions out for a given \(n\), but it’s also useful to have a systematic way to generate them and actually check that you have them all.
There is a recurrence relation \begin{align*} P_k(n) = P_k(n-k) + P_{k-1}(n-1) ,\end{align*} which follows from the fact that one can obtain a partition of \(n\) with \(k\) parts by either
- Taking a partition of \(n-k\) into \(k\) parts and adding 1 to each part, e.g. \([1,1,1,3] \mapsto [2,2,2,4]\)
- Taking a partition of \(n-1\) into \(k-1\) parts and adding a new standalone part \(1\), e.g. \([1,1,2,5] \mapsto [1,1,2,5,1]\).
Summing over \(k\) yields the following, which can be recursed: \begin{align*} P(n) &= \sum_{k=1}^n P_k(n-k) + P(n-1) \\ &= \sum_{k=1}^n P_k(n-k) + \sum_{k=1}^{n-1} P_k(n-1-k) + P(n-2) \\ &= \cdots ,\end{align*} where \(P_k(m) = 0\) for \(k>m\) and \(P_m(m) = 1\).
One can compute that \(P(5) = 7\), and the formula recovers this: \begin{align*} P(5) &= \sum_{k=1}^5 P_{k}(5-k) + P(4) \\ &= \qty{ P_1(4) + P_2(3) } + P(4) \\ &= \qty{ P_1(4) + P_2(3) } + \qty{P_1(3) + P_2(2)} + P(3) \\ &= \qty{ P_1(4) + P_2(3) } + \qty{P_1(3) + P_2(2)} + \qty{P_1(2)} + P(2) \\ &= \qty{ P_1(4) + P_2(3) } + \qty{P_1(3) + P_2(2)} + \qty{P_1(2)} + \qty{P_1(1) + P(1)} \\ &= \qty{1 + 1 } + \qty{1 + 1} + \qty{1} + \qty{1 + 1} \\ &= 7 .\end{align*} Note that you could just stop at the third line, since \(P(3) = 3\) is easy to enumerate: \([1,1,1], [1,2], [3]\).
Suppose \({\sharp}G = n = p^3 q^4\). Compute that \(p(3) = 3\) and \(p(4) = 5\), so there should be 15 abelian groups of this order. Enumerate the partitions:
- For 3: \([1,1,1], [1,2], [3]\)
- For 4: \([1,1,1,1], [1,2,1], [1,3], [2,2], [4]\)
Now for every distinct pair taking one from the first line and one from the second, we get a group of that order. A partition of \(m\) of the form \([a,b,c, \cdots]\) contributes a group of the form \({\mathbf{Z}}_{m^a} \times{\mathbf{Z}}_{m^b} \times{\mathbf{Z}}_{m^c} \cdots\).
Crossing \([1,1,1]\) with everything:
- \(\qty{{\mathbf{Z}}_p \times{\mathbf{Z}}_p \times{\mathbf{Z}}_p} \times\qty{{\mathbf{Z}}_q \times{\mathbf{Z}}_q \times{\mathbf{Z}}_q \times{\mathbf{Z}}_q} \mapsfrom [1,1,1] \times[1,1,1,1]\)
- \(\qty{{\mathbf{Z}}_p \times{\mathbf{Z}}_p \times{\mathbf{Z}}_p} \times\qty{{\mathbf{Z}}_q \times{\mathbf{Z}}_{q^2} \times {\mathbf{Z}}_q} \mapsfrom [1,1,1]\times[1,2,1]\)
- \(\qty{{\mathbf{Z}}_p \times{\mathbf{Z}}_p \times{\mathbf{Z}}_p} \times\qty{{\mathbf{Z}}_q \times{\mathbf{Z}}_{q^3}} \mapsfrom [1,1,1] \times[1,3]\)
- \(\qty{{\mathbf{Z}}_p \times{\mathbf{Z}}_p \times{\mathbf{Z}}_p} \times\qty{{\mathbf{Z}}_{q^2} \times{\mathbf{Z}}_{q^2}} \mapsfrom [1,1,1] \times[2,2]\)
- \(\qty{{\mathbf{Z}}_p \times{\mathbf{Z}}_p \times{\mathbf{Z}}_p} \times{\mathbf{Z}}_{q^4} \mapsfrom [1,1,1]\times[4]\)
Crossing \([1, 2]\) with everything:
- \(\qty{ {\mathbf{Z}}_p \times{\mathbf{Z}}_{p^2} } \times\qty{{\mathbf{Z}}_q \times{\mathbf{Z}}_q \times{\mathbf{Z}}_q \times{\mathbf{Z}}_q} \mapsfrom [1,2]\times[1,1,1,1]\)
- \(\qty{ {\mathbf{Z}}_p \times{\mathbf{Z}}_{p^2} } \times\qty{{\mathbf{Z}}_q \times{\mathbf{Z}}_{q^2} \times{\mathbf{Z}}_q} \mapsfrom [1,2] \times[1,2,1]\)
- \(\qty{ {\mathbf{Z}}_p \times{\mathbf{Z}}_{p^2} } \times\qty{{\mathbf{Z}}_q \times{\mathbf{Z}}_{q^3}} \mapsfrom [1,2] \times[1,3]\)
- \(\qty{ {\mathbf{Z}}_p \times{\mathbf{Z}}_{p^2} } \times\qty{{\mathbf{Z}}_{q^2} \times{\mathbf{Z}}_{q^2}} \mapsfrom [1,2] \times[2,2]\)
- \(\qty{ {\mathbf{Z}}_p \times{\mathbf{Z}}_{p^2} } \times{\mathbf{Z}}_{q^4} \mapsfrom [1, 2]\times[4]\)
And so on!
Classification: Groups of Special Orders
General strategy: find a normal subgroup (usually a Sylow) and use recognition of semidirect products.
- Keith Conrad: Classifying Groups of Order 12
- Order \(pqr\): ?
- Order \(p^2q\): ?
Every group \(G\) of prime order \(p\geq 2\) is cyclic and thus isomorphic to \({\mathbf{Z}}/p\).
Supposing that \(g\neq e\), it generates a cyclic subgroup \(H \coloneqq\left\langle{g}\right\rangle \leq G\) of order dividing \(p\) by Lagrange. Since \(g\neq e\), \({\sharp}H = p = {\sharp}G\).
Every group \(G\) of order \(p^2\) is abelian, and thus isomorphic to either \(C_{p^2}\) or \(C_p^2\).
Quotient by the center to get \(m\coloneqq{\sharp}G/Z(G) \in \left\{{ 1, p, p^2 }\right\}\). By cases:
- Since \(G\) is a \(p{\hbox{-}}\)group, \(G\) has nontrivial center, so \(m\neq 1\)
- If \(m=p\), then \(G/Z(G)\) is cyclic and thus \(G\) is abelian by the \(G/Z(G)\) theorem.
- If \(m=p^2\), \(Z(G) = G\) and \(G\) is abelian, done.
If \(G\) is a group of order \(pq\) where without loss of generality \(q<p\), then
- If \(q\notdivides p-1\) then \(G\) is cyclic and \(G\cong S_p \times S_q \cong C_{pq}\).
- If \(q\divides p-1\) then \(G\cong S_q \rtimes_\psi S_p\) where \(S_p {~\trianglelefteq~}G\) and \(\psi: S_q \to \mathop{\mathrm{Aut}}(S_p)\), and \(G\) has a presentation \begin{align*} G\cong \left\langle{a, b {~\mathrel{\Big\vert}~}a^p, b^q, bab^{-1}= a^\ell}\right\rangle \\ \\ \ell \not\equiv 1 \operatorname{mod}p && \ell^q \equiv 1 \operatorname{mod}p .\end{align*}
- Suppose \(q<p\).
-
Apply the Sylow theorems to \(p\):
- \(n_p \cong 1 \operatorname{mod}p \implies n_p \in \left\{{ 1, p+1, 2p+1, \cdots }\right\}\).
- \(n_p \divides q \implies n_p \in \left\{{ 1, q }\right\}\).
- Since \(1<q<p<p+1\), this forces \(n_p = 1\)
-
Suppose \(q\notdivides p-1\) and apply the Sylow theorems to \(q\):
- \(n_q \equiv 1\operatorname{mod}q \implies n_q \in \left\{{ 1, q+1, 2q+1,\cdots }\right\}\)
- \(n_q \divides p \implies n_q\in \left\{{ 1, p }\right\}\)
- Now note that if \(n_q\neq 1\), then \(n_q=p\) and \(p\) is of the form \(kq+1\) for some \(k\).
- Use of assumption: then \(p=kq+1 \iff p-1 = kq \iff q\divides p-1\), which is precisely what we assumed is not the case.
- So \(n_p = n_q = 1\) and \(S_p, S_q {~\trianglelefteq~}G\).
-
Apply recognition of direct products:
- \(S_p, S_q \leq G\): check.
- \(S_p, S_q {~\trianglelefteq~}G\): check.
- \(S_p \cap S_q = \left\{{ e }\right\}\): check, because they are coprime order.
- \(S_p S_q = G\): follows from a counting argument: \begin{align*} {\sharp}S_pS_q = {{\sharp}S_p {\sharp}S_q \over {\sharp}\qty{S_p \cap S_q}} = {pq \over 1} = {\sharp}G .\end{align*} If \(G\) is finite, then \(AB\leq G\) with \({\sharp}AB = {\sharp}G\) implies \(AB = G\).
- Suppose \(q \divides p-1\), the previous argument for \(S_p\) works, but the argument for \(S_q\) doesn’t, so we get a semidirect product.
- Work up to isomorphism: \begin{align*} S_p \cong {\mathbf{Z}}/p = \left\langle{a{~\mathrel{\Big\vert}~}a^p}\right\rangle &{~\trianglelefteq~}G \\ \\ S_q \cong {\mathbf{Z}}/q =\left\langle{b{~\mathrel{\Big\vert}~}b^q}\right\rangle &\leq G .\end{align*}
-
We have
\begin{align*}
G&\cong {\mathbf{Z}}/q \rtimes_{\psi} {\mathbf{Z}}/p
&& \psi: {\mathbf{Z}}/q \to \mathop{\mathrm{Aut}}\qty{{\mathbf{Z}}/p} \\ \\
\implies G &\cong\left\langle{a,b {~\mathrel{\Big\vert}~}a^p, b^q,\,\, aba^{-1}= \psi(b) = b^\ell }\right\rangle && \text{for some }\ell
.\end{align*}
- Since \({\mathbf{Z}}/q\) is cyclic, such a morphism is determined by the image of the generator \([1]_q \in {\mathbf{Z}}/q\).
- Note that \([1]_q \mapsto \operatorname{id}_{{\mathbf{Z}}/p}\) is such a morphism, and yields the direct product again.
- Identify \(\mathop{\mathrm{Aut}}({\mathbf{Z}}/p) \cong \qty{\qty{{\mathbf{Z}}/p}^{\times}, \times } \cong \qty{ {\mathbf{Z}}/(p-1), +}\).
-
So we need to classify morphisms
\begin{align*}
\psi: {\mathbf{Z}}/q\to {\mathbf{Z}}/(p-1)
.\end{align*}
- Consider \(\operatorname{im}\psi \leq {\mathbf{Z}}/(p-1)\).
- Sending \([1]_q\) to the identity in \(\mathop{\mathrm{Aut}}({\mathbf{Z}}/p)\) yields the direct product again, so pick nontrivial morphisms.
- Since \({\sharp}\operatorname{im}\psi \divides q\) which is prime, its order is equal to \(q\).
- Since \(q\divides p-1\) and \({\mathbf{Z}}/(p-1)\) is cyclic of order \(p-1\), by Cauchy’s theorem there is a unique subgroup of order \(q\), say \(C_q \leq {\mathbf{Z}}(p-1)\)
- We can send \([1]_q\) to \([\alpha]_{p-1} \in {\mathbf{Z}}/(p-1)\) where \(\alpha\) is any generator of \(C_q\), of which there are \(\phi(q) = q-1\) nontrivial choices.
- Thus there are \(q-1\) distinct nontrivial choices for the action \(\psi: {\mathbf{Z}}/q \to {\mathbf{Z}}/(p-1)\).
All choices yield isomorphic semidirect products.
-
Use that \(G\coloneqq A\rtimes_{\psi} N\) with \(\psi:A\to \mathop{\mathrm{Aut}}(N)\) is an \(\mathop{\mathrm{Aut}}(N)\) and \(\mathop{\mathrm{Aut}}(A)\) module, where \(f \in \mathop{\mathrm{Aut}}(N)\) and \(\pi\in \mathop{\mathrm{Aut}}(A)\) act in the following ways:
\begin{align*}
\pi &\curvearrowright A\rtimes_\psi N = A \rtimes_{\psi \circ \pi } N \\
f &\curvearrowright A\rtimes_\psi N = A \rtimes_{\gamma_f \circ \psi } N
.\end{align*}
where
\begin{align*}
\gamma_f: \mathop{\mathrm{Aut}}(N) &\to \mathop{\mathrm{Aut}}(N) \\
\psi &\mapsto f\circ \psi \circ f^{-1}
.\end{align*}
- These actions preserve the group isomorphism type of \(G\)
- However, since \(C_q \leq {\mathbf{Z}}/(p-1)\) and \(\mathop{\mathrm{Aut}}(C_q) \cong {\mathbf{Z}}/(q-1)\), there are exactly \(q-1\) automorphisms of the image \(C_q\), say \(\left\{{\pi_k}\right\}_{k=1}^{q-1}\).
- So \(\psi \circ \pi_k: {\mathbf{Z}}/q \to {\mathbf{Z}}/(p-1)\) for \(1\leq k \leq q-1\) yields \(q-1\) distinct actions, and we’re done.
If \(N{~\trianglelefteq~}G\) and \(P\in {\operatorname{Syl}}_p(H)\) then \(G = N_G(P)H\).
- Let \(g\in G\), then since \(P\leq H {~\trianglelefteq~}G\) we have \(gPg^{-1}\subseteq gHg^{-1}= H\).
- So \(P' \coloneqq gPg^{-1}\in {\operatorname{Syl}}_p(H)\) for all \(g\), and since Sylows in \(H\) are all conjugate, we can write \(P' = h^{-1}Ph^{-1}\) for some \(h\in H\).
- This says \(hPh^{-1}= gPg^{-1}\) and thus \(P = (g^{-1}h)P (h^{-1}g) = (h^{-1}g)^{-1}P (h^{-1}g)\).
- But then \(g^{-1}h \in N_G(P)\) so \(g\in N_G(P)H\).
Every finite \(p\) group is solvable.
- By induction on \(k\) in \({\sharp}G = p^k\): if \({\sharp}G = p\) then \(G\) is abelian and automatically solvable.
- Inductively, for \({\sharp}G = p^k\), now consider \(Z(G)\neq 1\) since we’re in a \(p{\hbox{-}}\)group.
-
If \(G/Z(G)\) is abelian, use the general fact: \(H\) solvable and \(G/H\) solvable implies \(G\) solvable.
- Here \(Z(G)\) and \(G/Z(G)\) are both abelian and thus solvable.
- Otherwise \(G/Z(G)\) is a \(p{\hbox{-}}\)group of size \(p^{k-1}\) and thus solvable by hypothesis.
If \({\sharp}G = pq\) with \(p<q\) distinct primes, then \(G\) has a normal subgroup of size \(q\).
This is immediate from Sylow theory: \([n_q]_q = 1, n_q \divides p, p<q\) forces \(n_q = 1\).
If \(|G| = pqr\) where \(p<q<r\) are distinct primes then \(G\) is solvable.
Idea:
- Get a normal subgroup \(R\) of order \(r\), so \({\sharp}(G/R) = pq\).
- Get a normal subgroup \(Q_1\) of order \(q\) in \(G/R\), which corresponds to \(Q{~\trianglelefteq~}G\) of order \(qr\) containing \(R\). Note that \(R{~\trianglelefteq~}Q\) since normality descends to subgroups.
-
Now \(G\to Q\to R \to 1\) is a subnormal series whose quotients are all cyclic and thus abelian:
- \({\sharp}(G/Q) = pqr/qr = p\),
- \({\sharp}(Q/R) = qr/r = q\),
- \({\sharp}(R/1) = r\),
Proof of first claim: let \(m\coloneqq{\sharp}G = pqr\), then \(G\) has a normal subgroup of order \(r\).
-
Claim: at least one of the Sylows for \(p,q,\) or \(r\) is normal.
-
If none of the Sylow \(p,q,r\) groups are normal, then \(n_r \geq r\) and \(n_p \geq q\). Counting the contributions from just \({\operatorname{Syl}}_q(G)\) and \({\operatorname{Syl}}_p(G)\) yields \begin{align*} n_q(q-1) + n_r(r-1) \geq pr(q-1) + pq(r-1) = pqr + p(qr -q - r) .\end{align*}
-
If this is to be at most \(m\), it must be that \(qr-q-r\) is negative (since \(p>1\) and otherwise this would yield more than \(pqr\) elements).
-
But if this holds, \begin{align*} qr-q-r \leq 0 \iff q(r-1) \leq r \iff q\leq {r\over r-1} .\end{align*} But \(q>2\) be assumption, and \(1\leq {r\over r-1}\leq 2\) for any number \(r\). \(\contradiction\).
-
So there is one of \(S_p , S_q, S_r\) that is normal in \(G\).
-
-
Now if \(S_r\) is normal we’re done, so suppose not and \(n_r > 1\). Claim: we can get another subgroup of order \(r\)
-
Let \(N\) be the normal Sylow, so either \(N\in {\operatorname{Syl}}_p(G)\) or \(N\in {\operatorname{Syl}}_q(G)\).
-
Then \(G/N\) has order \(r\ell\) for either \(\ell = q\) or \(\ell = p\) respectively.
-
In either case, \(\ell < r\). Using the lemma, \(G/N\) has a normal subgroup of size \(r\), say \(R/N \leq G/N\).
-
Then by the subgroup correspondence theorem, \(R\) corresponds to a normal subgroup \(R'{~\trianglelefteq~}G\) of size \(r\ell\) with \(r<\ell\).
-
Applying the same lemma to \(R'\) immediately yields a normal subgroup \(R''\) of order \(r\) in \(R'\)
-
Now use that \(R'' \operatorname{char} R'\) since Sylows are characteristic, and \(R'{~\trianglelefteq~}G\), so \(R''{~\trianglelefteq~}G\) too.
-