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Let $N$ be a normal subgroup of $G$ and $G$ a group. Let $H$ be a subgroup of $G$ with $|(N)||(H)|=|(G)|$ and $N\cap H=\{1\}$.

Why is $G=HN$ and why are two subgroups of $G$ isomorphic?

Thanks a lot!

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What do you mean by two subgroups of G are isomorphic? Do you mean N and H are isomorphic? This is not the case an $D_4=\{e,g,g^2,g^3,h,gh,g^2h,g^3h\}$ and then let $N=\{e,h\}$ and $H=\{e,g,g^2,g^3\}$ then $D_4=HN$ but H and N are not isomorphic? – hmmmm Oct 18 '12 at 17:51
@hmmmm if you take two subgroups of G, let's call them A and B, why are A and B isomorphic? – user33117 Oct 18 '12 at 17:58
but surely they don't have to be. You could take my example above, or $G$ and $\{e\}$ to be the subgroups and they are not isomorphic? – hmmmm Oct 18 '12 at 18:11
up vote 1 down vote accepted

There is a result in every basic group theory book (for example, Isaacs Algebra book 4.17) that $|HN|=\frac{|H||N|}{|H\cap N|}$ (at least, for finite groups).

In this case, that boils down to "$|HN|=|H||N|=|G|$". With $HN\subseteq G$, $|HN|=|G|$ implies $HN=G$.

The last part of your question is unclear, but there is something obvious that we can say which you might be looking for.

By an isomorphism theorem, $G/N=HN/N\cong H/(N\cap H)\cong H$.

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Perhaps the OP misread $G/N\cong H$ as "$N\cong H$"? – rschwieb Oct 18 '12 at 18:17
Maybe, i have asked in a comment – hmmmm Oct 18 '12 at 18:22

We have $|G| = |N||H|$ and $N \cap H$ is trivial. To see that $G=HN$, consider that the number of possible products of elements in $H$ and elements of $N$ is $|G|$. Suppose that $hn$ and $h'n'$ are two distinct products, and that $hn = h'n'$. Use the normality of $N$ to prove that $h=h'$ and $n=n'$. Then $NH$ has $|G|$ many elements, and hence must be the entire group.

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Hint: Consider $\phi: N \times H \to G$ given by $(n,h) \mapsto nh$. Prove that $\phi$ is injective. Conclude that it is surjective. Note that $\phi$ is not necessarily a homomorphism.

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I don't follow, if it is not necessarily a homomorphism then what is the point ? we already know there exist a bijection – Belgi Oct 18 '12 at 17:53
@Belgi the point is that it proves that $G=NH$, which is the image of $\phi$. – lhf Oct 18 '12 at 20:17

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