How many different groups of order $15$ there are?

Question

I wanted to share with you my resolution of this exercise.

How many different groups of order $15$ there are?

My resolution:

We're looking for groups such that $|G|=15=3\cdot 5$. Then: $G$ has an unique Sylow $3$-subgroup of order $3$ ($P_3$) and an unique Sylow $5$-subgroup of order $5$ ($P_5$).

Furthermore, $$|P_3 \cap P_5|\ \Large\mid\ \normalsize|P_3|,\ |P_5|$$ So $$|P_3 \cap P_5|=1\implies P_3 \cap P_5=\{e\}.$$

Then:$$|P_3\cdot P_5|=\displaystyle\frac{|P_3|\cdot |P_5|}{|P_3 \cap P_5|}=|P_3|\cdot |P_5|=3\cdot 5=|G|,$$

and $$G=P_3\cdot P_5 \simeq P_3 \times P_5 \simeq \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}.$$

It's everything correct and well justified? There's some things that I didn't say, for example that a subgroup a prime $p$ as order is cyclic, and thus isomoprhic to $\mathbb{Z}/p\mathbb{Z}$. Maybe I did something wrong. I would appreciate someone to look and criticize my exercise.

Thank you.

You didn't explain why there is only one Sylow $3$-subgroup. — Qudit, Mar 29 '15 at 17:19
You should be aware that $\Bbb Z/3\Bbb Z\times\Bbb Z/5\Bbb Z\cong\Bbb Z/15\Bbb Z$. — Tim Raczkowski, Mar 29 '15 at 17:21
@Qudit That's right, it's explained on the exercise before the posted one. We had to prove it generally for two $p$ and $q$. — Relure, Mar 29 '15 at 17:22
Alternatively, you might notice that the $15-5-3+1=8$ elements neither in $P_3$ nor $P_5$ have order dividing $15$, but neither dividing $3$ nor $5$, hence each of them generates $G$ as cyclic group. — Hagen von Eitzen, Mar 29 '15 at 17:40
The last line is not quite right. If $H$ and $K$ are subgroups of $G$ such that $G = H \cdot K$ and $H \cap K = {e}$, it doesn't follow that $G \cong H \times K$. It will be true when $H$ and $K$ are normal, though, so you can fix your proof by asserting that $P_3$ and $P_5$ are normal (being the unique group of their order in $G$). — Alex Zorn, Mar 29 '15 at 19:24
@AlexZorn Then, $G \cong H \times K$ it's only true when $H \cap K = {e}$, $G = H \cdot K$ and $H$ and $K$ are normal? — Relure, Mar 29 '15 at 19:28

stochasm · Answer 1 · 2015-03-30T01:59:29.053

Looks good, though you need to argue that one of $P_3$ or $P_5$ is normal in $G$ to know their product is a group (you have already shown this, as unique subgroups of a given order are normal). Here is a different way to do it:

As in your proof, let $P_3 \in \operatorname{Syl}_3(G)$ and $P_5 \in \operatorname{Syl}_5(G)$. As you say, it is easy to see these are unique (and thus normal). Now since $P_3$ has order $3$, $\operatorname{Aut}(P_3) \cong \mathbb{Z}/2\mathbb{Z}$. Since $N_G(P_3)/C_G(P_3)=G/C_G(P_3) \leq \operatorname{Aut}(P_3)$, and the order of $G/C_G(P_3)$ divides 15, it must be the case that $G/C_G(P_3)=1$. Thus, $C_G(P_3)=G$, so $P_3 \leq Z(G)$. Now take some nonidentity $x \in P_3$ and $y \in P_5$, noting that their order is 3 and 5, respectively. Since $x \in P_3 \leq Z(G)$, it follows that $x$ and $y$ commute, so the order of their product $xy$ is $\operatorname{lcm}(\lvert x \lvert,\lvert y \lvert)=15$. Thus, $G=\langle xy \rangle$, so $G$ is cyclic. It follows that every group of order 15 is cyclic, and hence isomorphic (in particular, they are isomorphic to the additive group $\mathbb{Z}/15\mathbb{Z}$).

The above method can be generalized to show that any group of order $pq$ for $p$ and $q$ distinct primes, $p<q$, and $p \nmid q-1$ is cyclic.

Perhaps a simpler argument, also using element orders, would be to note that since the $p$-Sylow subgroups are unique, there are $4$ elements of order $5$, $2$ elements of order $3$, and of course, a single element of order $1$ (the identity). Since $4+2+1=7<15$, there exists $x\in G$ of a different order. Thus, the order of $x$ must be $15$, and therefore $G\cong \left<x\right>\cong \mathbb{Z}/15\mathbb{Z}$. — doodle, Jun 24 '15 at 07:01
Ah, I see now that's exactly what @Hagen von Eitzen proposed in his comment. — doodle, Jun 24 '15 at 07:05

How many different groups of order $15$ there are?

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