Let G G be a homomorphism of a group with a group Prove th

Let : G G\' be a homomorphism of a group <G, > with a group <G\'\' , >. Prove that (a) if H\' is a subgroup of G\' , then ^-1 [H\' ] is a subgroup of G, (b) if the subgroup H\' of G\' is normal, then the subgroup ^-1 [H\' ] of G is also normal.

Solution

: G G\' be a homomorphism of a group <G, > with a group <G’, >

(a) H\' is a subgroup of G\'. To prove that ^-1 [H\' ] is a subgroup of G

Proof:

: G G\' be a homomorphism and H\' is a subgroup (and therefore a subset) of G\'. Hence, ^-1 [H\' ] is a subset of G.

Let a, b ^-1 [H\' ]

Then, there exists h, k H\' such that (a) = h, (b) = k

Now, (ab) = (a)(b)   because : G G\' be a homomorphism and hence operation preserving

                        = hk H\'        because H\' is a subgroup of G\'

This implies that ab ^-1 [H\' ]

Now, a ^-1 [H\' ] and ^-1 [H\' ] is a subset of G. This implies that a G

Since, G be a group, a^-1 exists, a^-1 G

Since, a ^-1 [H\' ] there exists h H\' such that (a) = h

Since, : G G\' be a homomorphism, (a^-1) = ((a))^-1

Therefore, (a^-1) = h^-1 H\'                because H\' is a subgroup of G\'

This implies that, a^{-1 -1} [H\' ]

For, a, b ^-1 [H\' ], we get ab ^-1 [H\' ] and a^{-1 -1} [H\' ]

This implies that ^-1 [H\' ] is a subgroup of G. (Proved)

(b) The subgroup H\' of G\' is normal. To prove that the subgroup ^-1 [H\' ] of G is also normal

Proof:

Let, g G and h ^-1 [H\' ]

If we can prove that ghg^{-1 -1} [H\' ] we are done.

Since, h ^-1 [H\' ], h G

Since, G be a group and g, h G, ghg^-1 G

(ghg^-1) = (g)(h)(g^-1)   because : G G\' be a homomorphism and hence operation preserving

Again, (g^-1) = ((g))^-1   because : G G\' be a homomorphism

Therefore, (ghg^-1) = (g)(h)((g))^-1  

Now, g G and h ^-1 [H\' ] implies that (g) G\' and (h) H\'

Since, The subgroup H\' is a normal subgroup of G\', (g)(h)((g))^-1 H\'

So, (ghg^-1) H\'

Therefore, ghg^{-1 -1} [H\' ]

This completes the proof.