Sylow-$3$-subgroups of group of order $117 = 3^2 \cdot 13$

Question

The possible orders of Sylow 3 subgroups is $\{1, 13\}$ (if I understood correctly). But how can I check the exact number? And how am I supposed to show that $S_3 = \Bbb Z_9$ or $S_3 = \Bbb Z_3\times \Bbb Z_3$?

Answer 1

Let me adress the latter question:

Every group of order $p^2$, where $p$ is some prime number, is isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$.

How can we see that?

Step 1: Show that if $G/Z(G)$ is cyclic, then $G$ is abelian.

Let $\langle \overline{a} \rangle = G/Z(G)$. That means that for $g,h \in G$ we can find integers $m$ and $n$ such that $\overline{a}^m = \overline{g}$ and $\overline{a}^n = \overline{h}$, i.e. there are $z,z' \in Z(G)$ such that we have $g = a^mz$ and $h = a^nz'$. We get $$gh = a^mza^nz' = a^ma^nzz' = a^{m+n}zz' = a^{n+m}zz' = a^na^mzz' = a^nz'a^mz = hg$$ as $z,z' \in Z(G)$. Thus $G$ is abelian.

Step 2: Take some group $G$ of order $p^2$. Then the center $Z(G)$ has order $1$, $p$, or $p^2$ by Lagrange's theorem. As $p$-groups have non-trivial center, only $p$ or $p^2$ can occur.

If $\text{ord}(Z(G)) = p^2$, then $G = Z(G)$ is abelian.

If $\text{ord}(Z(G)) = p$, then $G/Z(G)$ has order $p$ and thus $G/Z(G)$ is cyclic, such that $G$ is abelian (This also shows that this case actually cannot occur since $G$ being abelian contradicts that the quotient by the center is non-trivial).

Either way we established that $G$ is abelian such that we only have the $2$ isomorphism classes $\mathbb{Z}/p^2\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$ by the classification of finitely generated abelian groups.

Your first question:

I cannot think of an elementary argument right now, but there are $4$ isomorphism classes of groups of order $p^2q$ such that $p \mid q-1$, $p^2 \not\mid q-1$ and $q \not\mid p^2 -1$, namely:

1) The cyclic group of order $p^2q$.

2) The direct product of a cyclic group of order $pq$ and a cyclci group of order $q$.

3) The semidirect product of a cyclic group of order $q$ by a cyclic group of order $p^2$, where the action by conjugation of a generator is an automorphism of order $p$.

4) The semidirect product of a cyclic group of order $pq$ by a cyclic group of order $p$, where the action by conjugation of a generator is an automorphism of order $p$.

You could bruteforce your way through. What exactly is your task or the problem you want to deal with? Do you really need to find the number of sylow-$3$-subgroups? Maybe we can find some other argument if you want to show soemthing else.

Answer 2

You seem a bit confused. You are asking about the possible orders of the Sylow $3$-subgroups of $G$. Do you know how a Sylow subgroup is defined? If no, then in a group of order $117$ a Sylow $3$-subgroup has order $3^2$ and I recommend going back to your textbook or notes.

If yes and you really meant to ask how many Sylow $3$-subgroups $G$ has, then both possibilities occur. If $G$ is abelian, it has just one Sylow $3$-subgroup. If not (and this is possible), then it has $13$ Sylow $3$-subgroups.