Let G be a finite group with normal subgroup H a Let x eleme

Let G be a finite group with normal subgroup H. (a) Let x elementof G. Prove that |x H| (the order of xH as an element of G/H) divides |x| (the order of x as an element of G). (b) Let x elementof G. Prove that |xH| divides gcd([G: H], |x|). (c) Let x elementof G be such that |x| is relatively prime to [G: H]. Prove that x elementof H.

Solution

6(a) Let G be a finite group and H a normal subgroup of G.

To show that the order of [xH] in G/H divides the order of x in G.

Let the order of x in G be n.and the order of [xH] be k.

Consider [xH]ⁿ = xⁿ H =H. This means k divides n.

(b) Let x be in G. Show that order [xH] divides gcd ([G:H],[x]).

[xH] is an element of the group G/H , which has order [G:H].

Now, the order of any element in a finite group divides the order of the group

This implies order of [xH] divides [G:H]. But we already know from (a) that order of [xH] divides the

order of x in G.

Combining we get order of [xH] divides GCD ([G:H], order(x)) , as required.

(c) Let x in G be such that order of x is relatively prime to [G:H]. Prove that x is in H.

This is immediate from (b) as GCD ([G:H], order of x]=1, which implies order of [xH] is 1 ,

which is possible iff xH =H or x is in H.