Prove $G \cong H \times K$ when $H,K \lhd G$ and the orders of $H,K$ relatively prime with the product of their orders equaling the order of $G$

Question

Problem Statement

Suppose $G$ is a finite group. $H,K \lhd G$ normal subgroups, $\gcd(\lvert H \rvert, \lvert K \rvert)=1$ and $\lvert G \rvert = \lvert H\rvert \lvert K \rvert.$ Prove $G \cong H \times K$.

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My Approach

For this problem I intend to appeal to the following proposition:

Suppose that $H,K \lhd G$ are normal subgroups, $H \cap K = \{e\}$. Then the function $\phi: H \times K \to HK$ defined by $\phi(h,k) = hk$ is an isomorphism. In particular, if $HK = G$ then $G \cong H \times K$.

To that end, I endeavor to show, from the assumptions given:

1. $H \cap K = \{e\}$
2. $G = HK$

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My Attempt

For $1.$:

• My first thought was to utilize Lagrange's Theorem to say that since the orders of $H$ and $K$ are relatively prime , neither divides the other and so we can say that neither group is a subgroup of the other. Now this, to my understanding, doesn't mean that each subgroup has completely distinct elements (aside from the identity $e$, of course).
• To try and show that the elements must be distinct I thought about using a corollary to Lagrange's Theorem that says that "Suppose $G$ is a finite group and $g \in G$. Then $\lvert g \rvert \mid \lvert G \rvert$." To which we could then say that since the elements $h \in H$ must divide the order of $H$ they cannot divide the order of $K$ and vice versa and so the elements must be distinct in each subgroup since their orders are distinct. But since each is a subgroup of $G$ they must include the identity element.
  • For this, my understanding is still a bit limited. Does the corollary say that the order of ALL elements $g \in G$ must divide the order of $G$? It seems that way. I've only ever seen the order of an element be described in terms of the order of the cyclic subgroup generated by that element. But not every element generates a cylic subgroup, correct? So would this corollary still apply to the elements that do not generate a cylic subgroup? And what would that look like?

For $2.$:

• For this part I was thinking that perhaps the fact that $H \cap K = \{e\}$ (once proven from our initial assumptions) along with the fact that $\lvert G \rvert = \lvert H\rvert\lvert K\rvert$ would imply that each subgroup has distinct elements, and since the product of their orders is the order of the group $G$ then their product $HK$ would contain the same number of elements as $G$. Sort of like a partition where the cells are pairwise disjoint but their union is the whole partition.
  • However, this would imply that $\vert H\rvert\lvert K\rvert = \lvert HK\rvert$ and I don't believe this is generally true.

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Questions

First and foremost, is my approach correct. To appeal to this proposition and try to show that the assumptions we are given imply the $2$ other assumptions in the proposition?

Next, just my concerns listed in the "My Attempt" section. Namely: Whether I can use Lagrange's Theorem and the relevant corollary to show that the intersection of the two subgroups is simply the identity and the exact nature of the corollary. Also, if I'm on the right track thinking about how to show $G = HK$.

Answer 1

First, I think you're doing pretty well.

Next, I think you can do part one of your approach with Lagrange.

To wit, if $g\in H\cap K$, then $|g|\mid |H|$ and $|g|\mid |K|$. Thus $|g|=1$ and $g=e$.

For part two, I would appeal to the second isomorphism theorem. We get easily that $HK$ is a subgroup of order $|H||K|=|G|$. (Because the theorem says that $HK$ is a subgroup and that $HK/H\cong K$.)

And I believe that should wrap it up.