Please see attached image.
Is this map an isomorphism?
I understand that I have to show its bijective and a homomorphism but unsure whether these exist for the function. I would appreciate any help. Thanks
- DavidLv 41 month agoFavorite Answer
To show (or not) it is bijective, you need to show it is injective and surjective.
Pick any two elements of G, and show that θ(A) = θ(B) implies that A = B. This proves θ is injective. Since θ is just the inverse, this is quite simple (just utilise the definition of inverse, AA⁻¹ = I)
To show it is surjective, pick an element A in the range (which is just G again). Then just find a B in G such that θ(B) = A. This is simple again, using the definition of inverse.
Alternatively, you could just show that the inverse map exists. A map F is bijective if and only if it has an inverse. Since θ is just the inverse, the map is in fact self-inverse.
You now want to show (or not) that θ is a homomorphism. To do this, utilise the definition of a homomorphism, θ(A•B) = θ(A)•θ(B) for all A, B in G.
Then just plug in the definition of θ and see what happens.