Show that if $G$ is a finite group of order $n$, and $H$ is a subgroup of order $\frac{n}{2}$, then $H$ is a normal subgroup of $G$.
Please help me on this. I only know that $|gH| = |H|$. How can go on beyond that to attempt the problem?
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There is a nice generalization of this question: if p is the smallest prime numer dividing the order of the group, then any subgroup of index p is normal. – Sven Wirsing Oct 10 '16 at 08:06
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You can check this link to solve this problem – Truong Dec 26 '13 at 11:23
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Hints: What are the left cosets of $H$ in $G$? What are the right cosets of $H$ in $G$? Finally, note that if $g \in G$ but $g \notin H$, then $gH \neq H$ and $Hg \neq H$.
Michael Albanese
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1:To me the problem seems to be more general,please give me more direction. – nzobo Dec 26 '13 at 11:24
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What do you mean 'the problem seems to be more general'? Can you answer the questions I posed? – Michael Albanese Dec 26 '13 at 11:25
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2@nzobo: No more detailed information is needed here. Just reflect the directions in the post. +1 – Mikasa Dec 26 '13 at 11:35
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@nzobo: The main point is there are only two left or right cosets $H$ and $gH$ for any $g\in G\H$. Actually $gH=G\H$ for any $g\in G\H$ – Babai Dec 26 '13 at 11:35
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So, you want to prove $H\unlhd G $ i.e.,
$gH=Hg$ for all $g\in G$
Take some $g\in G$...
If your choice is from $H$ you have nothing to prove (??)
If your choice is not from $H$ then you can not immediately say $gH=Hg$
By $H$ is a subgroup of $G$ of order $\frac{n}{2}$ you actually mean there are only two cosets of $H$ in $G$..
Now Any two cosets are either equal or disjoint.. Why???
If $g\notin H$ then...
Can $H=gH$??
Can $H=Hg$??
If there are only two cosets...
how do you see containment of $G$ and $ H\cup gH$
how do you see containment of $G$ and $ H\cup Hg$