Showing H is a normal subgroup of G [duplicate]

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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marked as duplicate by YACP, Davide Giraudo, Sami Ben Romdhane, some1.new4u, ShuchangDec 26 '13 at 13:16

You can check this link to solve this problem

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I think this should have been as a good comment. :) –  Babak S. Dec 26 '13 at 11:33
You are welcome –  chuyenvien94 Dec 26 '13 at 11:54

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 $Hg \neq H$.

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:To me the problem seems to be more general,please give me more direction. –  nzobo Dec 26 '13 at 11:24
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
@nzobo: No more detailed information is needed here. Just reflect the directions in the post. +1 –  Babak S. Dec 26 '13 at 11:35
@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 $g\in G\H$ –  Susobhan Dec 26 '13 at 11:35

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 $H\cup Hg$

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