1

I want to prove that a group $G$ of order $15$ is isomorphic to $G_1\times G_2$ where $|G_1|=3$ and $|G_2|=5$.

To prove this I want to use the theorem that says that $G_1,G_2$ has to be normal subgroups,

$G_1 G_2=G$ and $G_1 \cap G_2 = \{e\}$.

To prove that $G_1,G_2$ are normal, I think I can use the fact that in general, a Sylow $p$-subgroup is normal if and only if $|\mathrm{Syl}_p(G)|=1$ and here $|\mathrm{Syl}_3(G)|=1$ and $|\mathrm{Syl}_5(G)|=1$ so $G_1,G_2$ are normal.

But I am stuck on how to prove the other two requirements. Suppose that $e\neq g \in G_1 \cap G_2$. I don't see how I can follow that and end up with a contradiction

Any hints on how to prove the last two ?

user26857
  • 52,094
領域展開
  • 2,139
  • 1
    What are the possible values of $|g|$? – Ayman Hourieh Jul 16 '21 at 11:53
  • @AymanHourieh The group $ G_1$ has $3$ elements and the group $G_2$ has $5$, so the possible elements of the intersection is $1,2,3$ if there is only one element it has to be $e$ if there is $3$ that means $G_1$ is a subgroup of $G_2$, but that would mean $3|5$, so the possible values are $1,2$ – 領域展開 Jul 16 '21 at 12:07
  • 1
    @PetrosK Are you saying that $2\mid3$ and $2\mid 5$? – Jyrki Lahtonen Jul 16 '21 at 12:08
  • @JyrkiLahtonen Why the numbers of elements of the intersection has to divide the orders of the groups ?, I don't see that, $G_1$ has $3$ elements which means it has also $2$ but $2$ cant divide $3$ – 領域展開 Jul 16 '21 at 12:11
  • 1
    $G_1$ has a subset with $2$ elements but no subgroup with $2$ elements. Do you know Lagrange's theorem? – paul blart math cop Jul 16 '21 at 12:13
  • 1
    @paulblartmathcop Oh, the intersection is also a subgroup, right ? So the only possible outcome is $1$ because the orders are relatively prime, correct ? – 領域展開 Jul 16 '21 at 12:16
  • Precisely! Just in case, make sure you can prove that the intersection is a subgroup. – paul blart math cop Jul 16 '21 at 12:18
  • @paulblartmathcop How about $G_1G_2=G$, can I say that $|G_1G_2|=|G_1||G_2|$ because the intersection is trivial ? Does this prove $G_1G_2=G$ ? – 領域展開 Jul 16 '21 at 12:19
  • That is the argument I would go for, but you should also make sure you know why $|G_1 G_2| = |G_1||G_2|$. The general fact is that $|G_1G_2| = |G_1||G_2| / |G_1 \cap G_2|$. Can you prove this? Maybe just in this specific case? – paul blart math cop Jul 16 '21 at 12:23
  • @paulblartmathcop I'm thinking that we divide with the intersection to prevent double counting, I will look it more later. – 領域展開 Jul 16 '21 at 12:47
  • Thanks everyone for helping. – 領域展開 Jul 16 '21 at 12:48

1 Answers1

1

$G_1G_2$ is a subgroup of $G$, and since $G_1$ and $G_2$ are subgroups of $G_1G_2$ we have that 3 and 5 divide the order of $G_1G_2$, so 15 divides the order of $G_1G_2$. It follows that the order of $G_1G_2$ equals 15 and $G_1G_2=G$.

On the other side, $G_1\cap G_2$ is a subgroup of $G_1$ and $G_2$, so its order divides 3 and 5, hence it is 1. It follows that $G_1\cap G_2=\{e\}$.

user26857
  • 52,094