Isomorphism check

[patter2809]

Q. Prove that the homomorphism, phi: G -> G', is an injective mapping if and only if Ker(phi) = { e }, where e is the identity in G.

A. Required to show: Phi is injective => Ker(phi) = {e} and Ker(phi) = {e} => Phi is injective.

For phi to be injective, a =/= b => phi(a) =/= phi(b). So, there can only be one g in G such that phi(g) = g' where g' is any element in G' which means the kernel can have only one element in it.

Furthermore, we have phi(e) = e' so e is certainly in ker(phi), and since |ker(phi)| = 1, ker(phi) = {e}.

For the second part, assume the contrary, i.e. Phi is not injective, so for some a,b in G we have phi(a) = phi(b).

phi(a) = phi(a*e) = phi(a) *' phi(e) = phi(a) *' e'
phi(b) = phi(b*e) = phi(b) *' e'

phi(a) *' e' = phi(b) *' e'
[phi(a)^-1 *' phi(a)] *' e' = [phi(a)^-1 *' phi(b)] *' e'
e' = phi(a)^-1 *' phi(b)
phi(e) = phi(a^-1 * b)

So, e = a^-1 * b
a * e = (a*a^-1) * b
a = b

Contradiction, so Ker(phi) = {e} => Phi is injective.

Is this rigorous, and if so, is there an easier way? Thanks in advance.

 

[daon2]

Q. Prove that the homomorphism, phi: G -> G', is an injective mapping if and only if Ker(phi) = { e }, where e is the identity in G.

A. Required to show: Phi is injective => Ker(phi) = {e} and Ker(phi) = {e} => Phi is injective.

For phi to be injective, a =/= b => phi(a) =/= phi(b). So, there can only be one g in G such that phi(g) = g' where g' is any element in G' which means the kernel can have only one element in it.

Not any g' in G'. That would imply \(\displaystyle \phi\) is surjective. A counter example is any inclusion map of a subgroup into the group it resides in.

Furthermore, we have phi(e) = e' so e is certainly in ker(phi), and since |ker(phi)| = 1, ker(phi) = {e}.

For the second part, assume the contrary, i.e. Phi is not injective, so for some a,b in G we have phi(a) = phi(b).

phi(a) = phi(a*e) = phi(a) *' phi(e) = phi(a) *' e'
phi(b) = phi(b*e) = phi(b) *' e'

phi(a) *' e' = phi(b) *' e'
[phi(a)^-1 *' phi(a)] *' e' = [phi(a)^-1 *' phi(b)] *' e'
e' = phi(a)^-1 *' phi(b)
phi(e) = phi(a^-1 * b)

So, e = a^-1 * b
a * e = (a*a^-1) * b
a = b

Contradiction, so Ker(phi) = {e} => Phi is injective.

Is this rigorous, and if so, is there an easier way? Thanks in advance.

Looks good. You could clean it up a little, there is much redundancy, and there was no need to assume phi is not injective (you did a direct proof).

\(\displaystyle \phi(a)=\phi(b) \implies \phi(a)^{-1}\phi(a)=\phi(a)^{-1}\phi(b) \implies e' = \phi(a^{-1}b) \implies e=a^{-1}b\implies\cdots\)

 

[patter2809]

Yeah, the phrasing on the first half is sloppy. Thank you!