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There is a group $G$ and subgroup $H \subset G$. Their orders:

$$\left | H \right| = n, \left |G \right| = 2n. $$

How can I prove that $H$ is a normal subgroup?

flipback
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    Hint: There are two cosets to examine. Normality means that $xH = Hx$ for all $x$... (btw you mean the group is twice as big as the subgroup in your title) – Elle Najt Apr 06 '16 at 15:45
  • @AreaMan I supposed that there is a subgroup {I, x} and I have to prove $xH = Hx$, but don't know how. – flipback Apr 06 '16 at 15:51

2 Answers2

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$G$ acts on the two element set $\{H, G\setminus H\}$ by multiplication from the left, thus giving rise to a homomorphism $G\to S_2$. What is the kernel?

Hagen von Eitzen
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  • I see the kernel should be H and it's enough to say that it 's normal. But I didn't catch how you got $ \left { H, G \setminus H \right }$. Could you give me a hint, please? – flipback Apr 06 '16 at 16:07
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Let $H$ be a subgroup of $G$ such that $|H|=n$ and $|G|=2n$. Then, we need only show that $xH=Hx$ for all $x \in G$. Let $x \in G$, then either $x \in H$ or $x \notin H$. If $x \in H$, then $xH = H = Hx$. If $x \notin H$, then since there are only two cosets (either left or right) and the one coset is always $H$, we must have $xH=Hx$.

sqtrat
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