$f : G → G′$ is a group homomorphism. (i). $H′$ is a normal subgroup of $G′$. Prove $f^{-1}(H′) = [ h ∈ G|f(h) ∈ H′] $ is a normal subgroup of $G$.
I don't know how to approach this question. Here is what I have so far:
Given $f$ is a group homomorphism, we have: for all $a, b ∈ G, f(a*b) = f(a) * f(b)$ as well as the inverse and identity properties associated with group homomorphism.
Given $H′$ is a normal subgroup of $G′$, we have that for all $h∈H′$, $g∈G′$: $ghg^{-1} ∈ H′$.
I know that we need to show that $f^{-1}$ and $G$ need to meet the above structure in order to prove a normal subgroup but I am now a little intimidated by all the notations and do not know how to proceed. Could anyone give me a hint?
(ii). If $H$ is a normal subgroup of $G$, is $f(H) = [f(h) ∈ G′|h ∈ H]$ always a normal subgroup of $G′$? If yes, prove it; if no, give a counterexample.
Could anyone also help me with part ii?
Thank you!
To see that $L = f^{-1}(H')$ is a subgroup of $G$, first note that it is a subset of $G$, because $G$ is the domain of $f$. Therefore, there is a binary operation $*$ on $L$, which is the restriction of $*$ on $G$, which is associative. All we need to show is that $1_G \in L$ (identity), $a,b \in L \implies ab \in L$ (closure) and $a \in L \implies a^{-1} \in L$ (inverse). This will show that $L$ is a subgroup.
To see that $1_G \in L$, note that $1_{G'} \in H'$, therefore $1_G \in f^{-1}(H') = L$ by the fact that $f(1_G) = 1_{G'}$, and the definition of $f^{-1}$.
Next, if $a,b \in L$, then $f(a) , f(b) \in H'$. Since $H'$ is closed under multiplication, $f(ab) = f(a)f(b) \in H'$, so $ab \in L$.
Can you show that $a^{-1} \in L$, using the fact that $H'$ is closed under inverses?
Now, if $H'$ is normal, then we want to show that $L$ is normal.
To show this, look at the definition. We want to show that for every $l \in L$, $g \in G$ we have $glg^{-1} \in L$.
Let us use the definition of $L$. It is by definition $f^{-1}(H')$. So, to show that $glg^{-1} \in L$, it suffices to show that $f(glg^{-1}) \in H'$.
Use the homomorphism property to get $f(glg^{-1}) = f(g)f(l)f(g^{-1}) = kf(l)k^{-1}$ where $k = f(g)$ is some element of $G'$. Now, can you use normality of $H'$ to conclude that $f(glg^{-1}) \in H'$? The key is that $f(l) \in H'$.
Hence, the preimage of a normal subgroup of the codomain group under homomorphism is always a normal subgroup of the domain group.
(An improvement of) This is often used as a tool along with Cayley's theorem etc. to show that groups of small sizes are not simple i.e. they contain non-trivial normal subgroups. You can see how : just find a homomorphism to another group, then a pullback of some normal subgroup of that group gives a candidate for a non-trivial normal subgroup of the group we are working with (though this method may not always work).
I request you to adopt a similar approach for part ii. Try to argue from definition of normal subgroup : if $g \in G'$ and $h \in f(H)$ then you want to show that $ghg^{-1} \in H'$. All you know is that $h = f(x)$ for some $x \in H$, but you cannot say the same of $g$. This inability to proceed should give you an idea of a counterexample.
For example, consider a group, a non-normal subgroup, and the inclusion map.