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I am trying to prove the following statement:

Let $G$ be a group and $H, K\mathrel{\unlhd}G$. Assuming that $H \cap K = \{1_G\}$ and $G = \langle H, K \rangle$, prove that $G \cong H \times K$.

From this, we get that $HK \cong H \times K$. Naturally, it would be ideal if I could prove that $HK \cong G$. One issue that I am facing here is the meaning of $G = \langle H, K \rangle$. How is this defined (I mean, more explicitly)? Is my strategy correct or

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2 Answers 2

up vote 3 down vote accepted

If both $H$ and $K$ are normal in $G$, note that $$\color{green}{\underbrace{\color{black}{h^{-1}k^{-1}h}}}\color{green}{\underbrace{\color{black}{k}}}\in \color{green}{K} \hspace{15pt}\text{and}\hspace{15pt}\color{blue}{\underbrace{\color{black}{h^{-1}}}}\color{blue}{\underbrace{\color{black}{k^{-1}hk}}}\in \color{blue}{H}$$ so we have $h^{-1}k^{-1}hk\in H\cap K=1$. Thus $h^{-1}k^{-1}hk=1$, so by left multiplying by $kh$ we get $hk=kh$. Since our choice was arbitrary, we must have that $H$ and $K$ commute elementwise.

$\langle H , K \rangle$ is the group generated by all words with letters in $H$ and $K$. Because $H$ and $K$ commute, however, we observe that given any such word, we can rearrange so that all $H$ are on the left and all $K$ are on the right. In particular, every element of $\langle H , K \rangle$ may be uniquely expressed as a product $hk$ for some $H\in H$ and some $k\in K$.

Now let $\varphi:\langle H , K \rangle\rightarrow H\times K$. Proving the homomorphic property is the content of this statement. Then, because elements of $G$ are uniquely represented as products $HK$, the isomorphism has a trivial kernel, so it's injective, so it's surjective, and we're finished.

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The symbol $\langle H, K \rangle$ is used to denote subgroup generated by elements of $H$ and elements of $K$. Since both $H$ and $K$ are normal subgroups, $HK$ is a subgroup. It should now be straightforward to prove that $HK = \langle H, K \rangle$

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