$3$-Sylows of $A_7$

Question

The problem states: "Find the number and the structure of the Sylow $3$-subgroups of $A_7$."

• Things I know:

-No problem with the structure, it is $C_3 \times C_3$ as they're no elements of order $9$ in $A_7$.

-From Sylow theorems we know that $n_3 \ | \ 280$ and $n_3 \equiv 1 $ mod $ (3)$.

• Best idea:

-The order of the conjugacy class of every $3$-cycle is $70$, so as all the Sylow are cojugated maybe we can do something with this, but I don't think I comprehend that part of the theory enough.

Answer 1

The group $A_7$ has order $7!/2 = 3^2\cdot(2^3\cdot 5\cdot 7)$. Recall for this special case the consequences of the three Sylow theorems:

Let $\Bbb S$ be the set of $3$-Sylow subgroups of $A_7$, then $\Bbb S$ is not empty, and in fact let us write explicitly one such group: $$ S=\Big\langle\ (123)\ ,\ (456)\ \Big\rangle\in\Bbb S\ , $$ each $H\in \Bbb S$ has $3^2$ elements, each two $3$-Sylow subgroups in $\Bbb S$ are conjugated, and denoting by $n=n_3$ the cardinality of $\Bbb S$ we have:

• $n$ divides the index $2^3\cdot 5\cdot 7$ of $S$ in $A_7$,
• $n$ is one modulo $3$,
• $n=|A_7:N|$, where $N$ is the normalizer of $S$ in $A_7$.

Let us use this information. What does a conjugated version of $S$ look like? It is also generated by two commuting cycles $s=(abc)$ and $t=(def)$. How many chances are there to choose them, so that different groups $H(s,t):=\langle s,t\rangle$ occur? It is clear that given such a Sylow group $H\in\Bbb S$ we can immediately isolate its two subgroups of $3$-cycles, so we can pick two generators $s$ and $t$, and there are some choices to do so, we can replace $s$ by $s^2$ and/or $t$ by $t^2$ and after these steps we may want to switch $s$ and $t$. Note that the set $\{a,b,c\}$ determines the subgroup $\langle s\rangle $ of $H(s,t)$. Similarly, the set $\{d,e,f\}$ determines the subgroup $\langle t\rangle $ of $H(s,t)$. Let $g$ be the remaining symbol.

Now we can count, there are $7$ choices for $g$, then $\binom 63$ choices for the set $\{a,b,c\}$, then the set of the remained symbols $\{d,e,f\}$ is determined, each such choice leads to a group $H(\ (abc)\ ,\ (def)\ )$, and when counting these groups, exchanging $\{a,b,c\}$ and $\{d,e,f\}$ leads to the same group. We have thus listed all conjugates of $S$; their number is $$ 7\cdot\binom 63\cdot\frac 12=7\cdot 20\cdot \frac12=70\ . $$ $\square$

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So the wanted number is $n=n_3=70$; as a check, it divides the index $2^3\cdot 5\cdot 7=280$, and it is $1$ modulo three.

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We are done, but let us give an alternative. Which is the normalizer $N$ of $S$ in $A_7$? First of all, the nine elements of $S$ are in $N$. Then $(23)(56)$ is also in $N$, conjugating with these elements brings the $S$-generators $s=(123)$ into $s^2=(132)$, and $t=(456)$ into $t^2=(465)$. And there is also $(14)(2536)$, also in $N$; we get it when trying to switch over $(123)$ and $(456)$. The first idea is to use $(14)(25)(36)$, which is not even, so we use an other odd permutation to adjust. So $N$ is a group with at least $36$ elements, which makes $|\Bbb S|=|A_7:N|$ have at most $7!/2/36=70$ elements. The divisors of $70$ which are one modulo three are $1,7,10,70$, and it is easy to eliminate the first three choices as being too small. In particular, the normalizer of $S$ is $$ N=\langle\ (123)\ ,\ (456)\ ,\ (23)(56)\ ,\ (14)(2536)\ \rangle\ . $$

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Note: I decided to submit the answer; from the comments, OP knows the structure $C_3\times C_3$ of each $3$-Sylow subgroup, to have the list of all such subgroups it is important to use the cycle structure. (Explicitly,
for $C_3\times C_3$ there is a lot of choices of two generators, but using also the cycle structure, we have less choices, and can count.)

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Note: A computer aided solution, here using sage:

sage: A7 = AlternatingGroup(7)
sage: S = A7.sylow_subgroup(3)
sage: N = A7.normalizer(S)
sage: A7.order() / N.order()
70
sage: S
Subgroup generated by [(4,5,6), (1,2,3)] of (Alternating group of order 7!/2 as a permutation group)
sage: N
Subgroup generated by [(4,5,6), (2,3)(4,6), (1,2,3), (1,4,3,6)(2,5)] of (Alternating group of order 7!/2 as a permutation group)

It may be useful to complement the theoretical part with explicit "dirty" experiences, for me this was working fine.