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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?

Shaun
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Nuke_Gunray
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    What about $G=\Bbb Z/4$ and $H=K=\Bbb Z/2$? Certainly $\Bbb Z/4$ is not isomorphic to $\Bbb Z/2\times \Bbb Z/2$. – Dietrich Burde Jan 31 '23 at 15:54
  • I'm quite sure that is a duplicate but, oddly enough, I can't find one; I tried on Approach0 and everything. I need to work on my web search skills. – Shaun Jan 31 '23 at 16:20
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    @Shaun Seems to be this (?) https://math.stackexchange.com/questions/541132/if-g-cong-h-k-does-it-follow-that-h-cong-g-times-k – Sine of the Time Jan 31 '23 at 16:22
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    I think the main source of confusion here is that $H$ is a quotient of $G$, not a subgroup of $G$. So you really have to say what you mean when you write something like $h = g+k$. I think if you define this clearly enough (e.g. $H$ is a normal subgroup of $G$ and the induced map $H\to G \to G/K\cong H$ is an isomorphism), then I think this is true. – Mathmo123 Jan 31 '23 at 16:29
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    @amWhy - I don't think this question is a duplicate of the one you've linked. It has an additional assumption that $H\lhd G$, so in particular, with suitable modifications, it is actually true! – Mathmo123 Jan 31 '23 at 16:39
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    It is perverse to write $HK$ (implying multiplicative notation) and then using $+$ for the group operation. When the group is multiplicative, $HK={hk\mid h\in H,k\in K}$. When the group is additive, the set ${h+k\mid h\in H, k\in K}$ is denote $H+K$, as one might expect. And it is very bad practice to use $+$ to denote an operation that need not be commutative, as is the case here. So your notation could be said to be rather sucky throughout. – Arturo Magidin Jan 31 '23 at 19:31
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    Let $G$ be cyclic of order $2$, and let $H=K$ be the subgroup of order $2$. Then $G/K\cong H$, but $G$ is not isomorphic to $H\times K$, which is the Klein $4$-group. – Arturo Magidin Jan 31 '23 at 19:33

2 Answers2

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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).

reuns
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  • Thank you for your answer! But doesn't $H \cong G/K$ already mean by definition that there exists an isomorphism from $H$ to $G/K$? – Nuke_Gunray Jan 31 '23 at 23:30
  • Sorry but how does it relate to my answer? And no $\cong$ is an extremely ambiguous symbol, it is supposed to mean "there exists an isomorphism" but in practice it usually means "the obvious/natural/canonical map is an isomorphism". – reuns Jan 31 '23 at 23:37
  • Oh okay, I see now what you actually said in your answer. Of course it is not enough to have just the existence of some isomorphism $-$ we explicitly want $h\mapsto hK$. Sorry for the unneccessary comment before :-) – Nuke_Gunray Jan 31 '23 at 23:40
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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.

Arturo Magidin
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