Given $G$ - finite group, $N$ - its normal subgroup, $|N|$ and $|G/N|$ are relatively prime. Prove that if $H$ - is subgroup of $G$ and its order is equal to order of $G/N$, then $HN = G$.
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Do you see that $N$ and $H$ intersect trivially? – Tobias Kildetoft Oct 17 '16 at 09:55
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Because of $|N|$ and $|G/N|$ are relatively prime? – Welez Oct 17 '16 at 09:56
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Yes. Now just apply the formula for the order of the product of subgroups. – Tobias Kildetoft Oct 17 '16 at 09:57
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As $N\lhd G$, $HN$ is a subgroup of $G$. We know that $H\le HN$ and $N\le HN$, hence $|H|$ and $|N|$ both divide $|HN|$.
Hagen von Eitzen
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Let $p:G\rightarrow G/N$ the quotient map, and $x\in H$, $p(x)=1$ implies that $x\in H\cap N$, this implies that $ord(x)||N|$, since $ord(x)| |H|$ and $gcd(|N|,|H|)=1$, we deduce that $x=1$. This implies that the restriction of $p$ to $H$ is onto since $|H|=|G/N|$.
Let $y\in G$, there exists $x\in H$ $p(x)=p(y), x(x^{-1}y)=y, x^{-1}y\in N$ since $p(x^{-1}y)=1$.
Tsemo Aristide
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Hint:
Prove that $H\cap N=\{e\}$. As a consequence $HN\simeq H\times N\;$ is a subgroup of $G$ such that $\lvert HN\rvert=\lvert G\rvert\;$.
Bernard
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@Tobias Kildetoft: It adds nothing to your comment. I guess either we posted at the same moment, or I didn't check all the comments. – Bernard Oct 17 '16 at 12:46
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@Bernard possibly a silly question, but why does $H \cap N = {e}$ imply that $HN \simeq H \times N$? I'm working on this problem here: http://math.stackexchange.com/questions/2102574/g-pq-where-p-and-q-are-primes-is-the-semidirect-product-of-subgroups-o Showing that $G = HN$ is essentially the key to my figuring it out. – Jan 18 '17 at 05:16
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@JessyunBourne: That is because each element of $G$ can be written as $hn$ ($h\in N, n\in N$) in a unique way. Indeed, if $hn=h'n'$, we deduce $h'^{-1}h=n'n^{-1}\in H\cap N$, hence $h'^{-1}h=n'n^{-1}=e$, i.e. $h=h,\enspace n=n'$. – Bernard Jan 18 '17 at 09:35