Does $G/K \cong H$ imply that $G \cong H\times K$ for normal $H,K$?

Question

I've seen a lot of posts here concerning the reverse statement, but I am wondering whether or not one has the following:

Let $H,K$ be normal subgroups of a group $G$, and assume that $G/K\cong H$. Does this suffice to say that \begin{equation}G \cong H\times K \tag{$*$}\end{equation} holds?

It looks a bit to easy to be true, but my reasoning is as follows: One can prove (see e.g. this question) that $(*) $ holds if $H,K$ satisfy

• $ HK := \{h+k : h\in H,k\in K\} =G$,
• $H\cap K = \{e\}$.

Now, from $G/K \cong H$, one has that for each $h\in H$, there exists exactly one $g\in G$, so that $h = g + k$ for some $k\in K$, or conversely that for each $g$, there exist $h,k$, so that $g = h-k\in HK$, which proves the first equality.

So it all boils down to whether or not the second equality can be shown. For finite groups, this should be trivial: One has $|G|/|K| = |H|$, so because of $|G| = \frac{|H|\cdot |K|}{|H\cap K|}$, we get that $|H\cap K| = 1$.

But what about more general groups?

Answer 1

If $H,K$ are two normal subgroups of $G$ such that the map $H\to G/K,h\mapsto hK$ is an isomorphism then obviously $H\cap K=\{1\}$ (the map is injective) and $G = \{ hk, h\in H, k\in K\}$ (the map is surjective).

It remains to check if $H,K$ commute!

$H$ is normal so $khk^{-1}$ is an element of $H$. So $khk^{-1} = h_2, kh = h_2k$.

$K$ is normal so $Kh = hK$ ie. $h_2k\in hK$ and $h_2K = hK$.

As $h\mapsto hK$ is injective it must be that $h_2=h$. Whence $h,k$ commute and we are done: $H\times K\to G, (h,k)\mapsto hk$ is an isomorphism (an homomorphism, injective and surjective).

Answer 2

The statement does not hold as written; it requires further assumptions, such as the ones given by reuns.

For a counterexample, consider $G=C_4$ the cyclic group of order $4$, and let $H$ and $K$ be equal to each other, the cyclic group of order $2$. Then $G/K\cong H$, but $H\times K$ is the Klein $4$-group, which is not isomorphic to $G$.

Examples with $H\neq K$ are easy to construct. Let $n$ be a positive integer that can be written as $n=abc^2$ with $a,b,c$ positive integers greater than $1$, $G=C_{n}$, take $K$ to be the unique subgroup of order $ac$ and $H$ the unique subgroup of order $bc$. Then $G/K$ is cyclic of order $bc$, hence isomorphic to $H$, but $G$ is not isomorphic to $K\times H$, as the latter contains a subgroup isomorphic to $C_c\times C_c$, which is not cyclic. Selecting $n$ with $a\neq b$ gives $K\neq H$ (and in fact $K\not\cong H$).

The error in your argument becomes clear if you actually keep track of the fact that you have isomophisms. Let $\phi\colon G/K\to H$ be the isomorphism, and let $\pi\colon G\to G/K$ be the canonical projection; and let $\iota\colon H\to G$ be the subgroup embedding.

You argue that for each $h\in H$ there exists $g\in G$ and $k\in K$ such that $h=gk$. But in fact what you have is that for each $h\in H$ there exists a unique $gK\in G/K$ such that $\phi(gK) = h$; this does not let you express $h$ as equal to $gk$ for some $k\in K$, though, because you are using $\phi$ to get to $H$. You would be able to conclude that if you also had that for each $h\in H$ you knew that $\phi(hK) = h$. If you had that additional assumption, then from $\phi(gK)=h=\phi(hK)$ you would be able to conclude that $gK=hK$ and therefore that $h=gk$ for some $k\in K$, as you want. But this additional condition is nowhere present in your listed hypotheses, so it cannot just be assumed.

In short, the first equality is not guaranteed by your hypotheses, so the conclusion does not follow.