4
$\begingroup$

I want to prove the following exercise.

If a group $G$ is the direct product of subgroups $H,K$, then $K$ is isomorphic to $G/H$.

To prove this, I think I need first to show $H$ is normal in $G$.

I can show that there is a normal subgroup $J$ in $G$ that is isomorphic to $H.$ But I don’t know how to show $H$ is a normal subgroup in $G.$

I’m not sure but my guessing is that $gHg^{-1} = J$ for some $g \in G$, so that $H = J.$ But . . . maybe there’s a counterexample.

Also, if I show $H$ is normal somehow, I still don’t know how the conclusion of the exercise follows from it. When $G = H\times K$ means the internal direct product of its normal subgroups, I can solve the exercise. If not, I know that, by using $\pi_k : H\times K \to K$ (the canonical projection), we can show $H\times K/\ker(\pi_k)$ is isomorphic to $K.$ But how can i show that $G/H$ is isomorphic to $H\times K/\ker(\pi_k)$?

Can somebody help?

Thank you!

$\endgroup$
9
  • 3
    $\begingroup$ This is actually a lot easier than what you are doing. Just let $G=H\times K$. Then, clearly $H,K$ are both normal in $G$, and are the quotients of the group modulo the other subgroup. The homomorphisms that induce these quotients are simply the forgetful morphisms. $\endgroup$ Commented Oct 25, 2019 at 14:26
  • $\begingroup$ I think we need to interpret $G = H$ x $K$ as $G$ is isomorphic to $H$ x $K$.. $\endgroup$
    – anadad
    Commented Oct 25, 2019 at 14:29
  • $\begingroup$ I don’t understand why H is normal in G clearly. $\endgroup$
    – anadad
    Commented Oct 25, 2019 at 14:30
  • $\begingroup$ Because of the forgetful morphism from $H\times K\to K$. $\endgroup$ Commented Oct 25, 2019 at 14:32
  • 2
    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Commented Oct 25, 2019 at 14:36

2 Answers 2

1
$\begingroup$

By definition, if $G$ is the internal direct product of $H$ and $K$, then $H$ and $K$ are both normal, $H\cap K=\{e\}$, and $HK=G$. There is no need to prove this because this is what you are given. Then you can use the relevant isomorphism theorem to show that $HK/H\simeq K/(K\cap H) $.

$\endgroup$
7
  • $\begingroup$ So, in the exercise, can I just assume that G is the internal direct product of H and K? $\endgroup$
    – anadad
    Commented Oct 25, 2019 at 14:37
  • $\begingroup$ @anadad If it is the direct product of subgroups, then it is the internal direct product. Even if it's external you can identify the subgroups with $H$ and $K$. $\endgroup$ Commented Oct 25, 2019 at 14:39
  • $\begingroup$ Thanks. So you mean that whenever I see $G = H \times K$ and $H,K$ are subgroups of G, I can assume $HK=G$, $H \cap K= e$ and they are normal in $G$ ? $\endgroup$
    – anadad
    Commented Oct 25, 2019 at 14:51
  • 3
    $\begingroup$ Here Matt is talking about the direct product up to isomorphism, technically speaking, @anadad; there is a distinction between internal and external direct products - so you're right - and showing that they are isomorphic is often left as an exercise. You can see the difference between them in Gallian's "Contemporary Abstract Algebra". $\endgroup$
    – Shaun
    Commented Oct 25, 2019 at 14:55
  • 2
    $\begingroup$ @Shaun thanks I will look it up! $\endgroup$
    – anadad
    Commented Oct 25, 2019 at 15:04
0
$\begingroup$

The group $G\cong H\times K$ with $H\cong \langle S_H\mid R_H\rangle$ and $K\cong \langle S_K\mid R_K\rangle$ has as a presentation $$G\cong \langle S_H\cup S_K\mid R_H\cup R_K\cup X\rangle,$$ where $X=\{hk=kh\mid h\in S_H\land k\in S_K\}$, from which it is easy to see that $H\cong L$ such that $L\unlhd G$. (Why?)

$\endgroup$
2
  • $\begingroup$ Thank you for your help, but I don’t know about the presentation. $\endgroup$
    – anadad
    Commented Oct 25, 2019 at 14:48
  • $\begingroup$ You're welcome, @anadad. Nevermind :) $\endgroup$
    – Shaun
    Commented Oct 25, 2019 at 14:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .