Abstract-H is normal to G. Order of G/H is m, g^m exists in H?

[danthaman857]

Let H be a normal subgroup of a finite group G. The order of G/H (quotient/factor group) is m. Show g^m is in H for all g in G.


Lagrange's Thm says that o(G) = o(H) * o(G/H)
xH = Hx for all x in G, since H is a normal subgroup of G

Ok, I've got a lot of statements that i believe to be true, how relevant they are is what i can't decide.

- let o(G) = k, o(H) = j, o(G/H) = m
so k = mj by Lagrange

- the quotient group is broken up into m disjoint partitions of G
- each partition has j elements
- I know that the o(g) | o(G)

 

[Office_Shredder]

If the order of G/H is m, then what is (g+H)^m?

 

[danthaman857]

If + is the Binary op, then (aH)^m should be equal to (aH)(aH)... m times and since H is normal in G aH = Ha so...

(aH)^m = a^mH?

so a^m is in G

 

[danthaman857]

if a^m was to equal the identity in G, then a^mH = H would imply that a^m is an element of H because hH = H = Hh, but i don't see how a^m could equal the identity in G