Identify the kernel and whether or not the following mapping is 1-1 or onto
1. G group, φ: G-> G defined by φ(a) = a^-1 for a is an element of G
How can I find out if a homomorphism is 1-1 or onto
Thanks
Identify the kernel and whether or not the following mapping is 1-1 or onto
1. G group, φ: G-> G defined by φ(a) = a^-1 for a is an element of G
How can I find out if a homomorphism is 1-1 or onto
Thanks
$\displaystyle ker \phi=\{x \in G:\phi(x)=e\}=\{e\}$. The second equality obtained from the definition of $\displaystyle \phi$
$\displaystyle ker \phi=\{e\}$ iff $\displaystyle \phi$ is one-to-one. However, you have to have the assumption that $\displaystyle \phi$ is a homomorphism.
1-1 and onto are two separate concepts, so there is no 'or' concept here.
for 1-1: Show if φ(g1) = φ(g2) then g1=g2. This is very easy in this problem
for onto: Show for ever g in G, there is an h in G such that φ(h) = g. This is again easy here as it follows directly from Group Axioms
I would also like to point out here that this mapping is not a homomorphism unless the group is abelian.
$\displaystyle \phi(ab)=(ab)^{-1}=b^{-1}a^{-1}$
$\displaystyle \phi(a)\phi(b)=a^{-1}b^{-1}$
So this is a homomophism if anf only if $\displaystyle a^{-1}b^{-1}=b^{-1}a^{-1}\Leftrightarrow aba^{-1}b^{-1}=1 \Leftrightarrow ab=ba$ for all $\displaystyle a,b \in G$. That is G is abelian.