Show that $a^m$ is in $H$ for every $a$ in $G$

Question

Let $H$ be a normal subgroup of $G$, and let $m=(G:H)$. Show that $a^m \in H$ for every $a \in G$.

I have been thinking about this question for a few days but I get something informal.

What am I missing?

There are $m$ $H$-cosets.
If we consider the cosets of $e$, $a$, $a^2$, $\cdots$, $a^{m-1}$, we get $m$ distinct cosets(actually I claim this without a valid proof, any hints?) Then the coset of $a^m$ should be one the above coset. (Then why must it be $eH$?)

Thanks in advance.

If a group $G$ has order $n$, and $a \in G$, then what can you say about $a^n$? — , Nov 13 '14 at 11:25
Your claim that the cosets $a^i H$ are distinct is not correct. — Martin Brandenburg, Nov 13 '14 at 11:25
See also http://math.stackexchange.com/questions/573050/if-h-is-a-subgroup-of-g-of-finite-index-n-then-under-what-condition-gn. — lhf, Nov 13 '14 at 11:33

score 1 · Accepted Answer · answered Nov 13 '14 at 11:33

1

let xH be any element of G/H, then (xH)^m=H as |G:H|=n this implies that x^nH=H then x^n belongs to H for all x in G.

answered Nov 13 '14 at 11:33

Yogi

716

score 0 · Answer 2 · answered Nov 13 '14 at 11:24

0

Hint.

Do you know the case $H=1$?

Does the general case reduce to this case?

answered Nov 13 '14 at 11:24

Martin Brandenburg

163,620

score 0 · Answer 3 · answered Nov 13 '14 at 11:27

0

Hint: look at $G/H$, being a group of order $m$. What do you know about orders of group elements and the order of the group they live in? Joseph-Louis Lagrange could help.

answered Nov 13 '14 at 11:27

Nicky Hekster

49,281

Show that $a^m$ is in $H$ for every $a$ in $G$

3 Answers3

Linked

Related