Question

Suppose H is a subset of G is a normal subgroup of index k. Prove that for any a in G, a to the power of k in H. Does this hold without the normality assumption?

Answer 1

Let H ≤ G is a normal subgroup of a finite group G of finite index finite k, i.e., |G| /|H| = k

Considering the canonical projection f :G→G/H. We know G/H is a finite group of order k by assumption. Thus (f(g))ⁿ= f(g^{​​​​​​n})=1 in G/H, i.e. back in G we have g^k ∈ H. Done.

In other way

Since |G/H| = n<∞ and G/H = {H, g^​​​​​​​​​​H, g²H, ..., g^n-1H}. Then for any gH in G/H

(gH)^k = g^kH =  H,

whichcompletes the proof.

Clearly without normality condition above condition doesn't hold.