Question

An isomorphism between groups G and G'' is just a bijective function f:G→G'' that respects the group operation. In other words, for all a, b in G, f(ab) = f(a) f(b) in G''. (a) Prove that if G and G'' are groups and f:G→G'' is an isomorphism, then f(e G) = eg'', where eG and eG'' are the identity elements of G and G'' respectively. (b) Prove that if G and G'' are groups and f:G→G'' is an isomorphism, thenthen for any element a in G, f(a^-1) = f(a))^-1 in G''. You can proceed with proving both parts (a) and (b) related to group isomorphisms.

Answer 1

Part (a) of the question can be proven by showing that f(eG) = eG''. We can use the properties of group isomorphisms to prove that f(a^-1) = (f(a))^-1 in G'' for part (b).

Step-by-step explanation:

Part (a)

To prove that f(e

G

) = e

G''

, we know that f is an isomorphism, so it is bijective. Therefore, there exists an inverse function f

-1

: G'' → G. Since f is a group isomorphism, it preserves the group operation, which means for any a,b ∈ G, f(ab) = f(a)f(b) in G''.

By the definition of the identity element, f(e

G

) = e

G''

implies that f(e

G

)f

-1

(e

G''

) = e

G''

. Since f is bijective, f

-1

(e

G''

) = e

G

. Therefore, we have f(e

G

)e

G

= e

G''

, which implies that f(e

G

) = e

G''

.

Part (b)

To prove that f(a

-1

) = (f(a))

-1

, we use a similar approach.

Once again, f is a group isomorphism, so for any a,b ∈ G, f(ab) = f(a)f(b) in G''.

Let x denote f(a

-1

) and y denote (f(a))

-1

. Then we have x = f(a

-1

) in G'' and y = (f(a))

-1

in G''.

Applying f to both sides of y = (f(a))

-1

, we get f(y) = f((f(a))

-1

) = f(f(a))

-1

in G''.

Since f is bijective, f(f(a)) = a and f is a group isomorphism. Therefore, f(y) = a

-1

in G.

Combining x = f(a

-1

) and f(y) = a

-1

, we have x = a

-1

in G and x = f(a

-1

) = (f(a))

-1

in G''.

Therefore, f(a

-1

) = (f(a))

-1

in G''.