Let $\phi:G\rightarrow G'$ be a group homomorphism and $H$ a subgroup of $G$ containing $\ker(\phi)$. Cont. below...

Question

Define the homomorphism $\psi:H\rightarrow G'$ by $\psi=\phi\circ i$ where $i:H\rightarrow G$ is the inclusion homomorphism $i(h)=h\space\space\forall \space h\in H$. Show that $\ker(\phi) = \ker(\psi)$

So I know that I somehow have to use the correspondence theorem to prove this, but the way my book describes the theorem, and the way the question is presented are very different. In my book, they tell us to let $g$ be the restriction of $\phi$ to some set $S^{\ast}=[x\in G:\phi(g)\in S]$ where $S$ is any subgroup of $H$. Not only does that not make a lick of sense to me, I can't see the correlation to the problem at hand. Any and all help will be appreciated. BTW I am very beginner at abstract algebra and my English isn't the best.

Answer 1

Not sure why such advanced theorems are needed here.

Suppose $x \in \ker (\phi)$. That means, $\phi(x) = 1_{G'}$ (the identity element in $G'$), and also that $x \in H$. Now let's examine $\psi(x)$: $$ \psi(x) = \phi \circ i (x) = \phi(x) = 1_{G'}, $$ so $\ker(\phi) \subset \ker(\psi)$.

Suppose now that $x \in \ker (\psi)$. It follows that $x \in H$ and that $$ 1_{G'} = \psi(x) = \phi \circ i (x) = \phi ( i (x) ) = \phi ( x ). $$ So, $\ker(\phi) \supset \ker(\psi)$, and we are done.

I am not even sure what the learning value of this exercise is for group theory.

To you as a beginner in abstract algebra, may I recommend two things:

• When faced with abstraction, try to come up with examples. For your problem, consider this example: $G$ = the multiplicative group of all invertible 2-by-2 real matrices, $G'$ = the multiplicative group of the nonzero reals, $\phi(x) = \det x$, and $H$ the group of all 2-by-2 real matrices with determinant $-1$ or $1$. How transparent is the statement you want to prove for this example?
• Use a good source on group theory, which shows how group theory is really geometric: Part I of Abel's theorem in problems and solutions. I have seen no equals in quality to this exposition.