I think I should state the original problem before I begin.
Let $a = (2, 4)$, $b = (1, 2, 3, 4)\in S_4$. (Symmetric group) Let $G = \left\langle a, b\right\rangle$.
Prove that $H =\{id, b^2\}$ is a normal subgroup of $G$, and $G/H$ is isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.
Motivation
It is pretty obvious that the permutations $a$ and $b$ describe the symmetries of a square, so $G$ is actually equal to the $4^{\text{th}}$ dihedral group $D_4$.
As I was solving this problem by actual calculations, I felt that something very important is going on hidden in the background. I thought that there must be an important fact in group theory, making this problem make sense, or perhaps, even generalize the problem (for dihedral groups).
So I felt the urge to use the First Isomorphism Theorem for this. I thought there might be a way to construct some surjective group homomorphism $\varphi: D_4 \rightarrow \mathbb{Z}_2\times\mathbb{Z}_2$, so that $H = \ker \varphi$, and the results would automatically follow.
The kernel of a group homomorphism is always a normal subgroup, $\ker\varphi \unlhd D_4$, and by the First Isomorphism Theorem, I would be able to directly derive that $\DeclareMathOperator{\im}{im}$ $G/H = D_4/\ker \varphi \approx \im \varphi = \mathbb{Z}_2\times\mathbb{Z}_2$.
So I tried to construct this surjective group homomorphism, but this didn't work well. So I kind of reverse engineered the process. I calculated $G/H$ and got $$G/H = \{\overline{id}, \overline{a\vphantom{b}}, \overline{b}, \overline{ab}\}$$ If some isomorphism $\overline{\varphi}: G/H \rightarrow \mathbb{Z}_2\times\mathbb{Z}_2$ actually existed by the First Isomorphism Theorem, it should be defined as the following: $$ \overline{\varphi}(\overline{g}) = \varphi(g) \qquad (g\in G)$$ Using this, I could actually try to construct the surjective group homomorphism that I wanted in the first place. Thus I tried the following $$ \begin{aligned}\varphi(a) &= (1, 2)(3, 4) = ba = ab^3 \\ \varphi(b) &= (1, 3)(2, 4) = b^2 \\ \varphi(ab) &=(1, 4)(2, 3) = ab\end{aligned} $$ and tried to find an explicit formula for the homomorphism. Something like $\varphi(x) = \;???$ for all $x\in D_4$. (For example, something like $\varphi(x) = ab^2x$) But I failed to find such a formula, even though I tried using the properties $a^2=id$, $b^4=id$, $ba=ab^3$.
I know that there are more settings (for the values of $\varphi(a), \varphi(b)$) so that $\varphi$ would be a homomorphism, but I am stuck without any luck.
By the way, the reason I'm trying to find an explicit formula is to prove that $\ker \varphi = H$, and that $\varphi$ is indeed a surjective group homomorphism. Maybe there is a way to get around this?
Summary
- Is there an explicit formula for a surjective group homomorphism $\varphi: D_4\rightarrow \mathbb{Z}_2\times\mathbb{Z}_2$?
- If not, why can't we find one? If there is a reason, I would really like to know about it. Is it because of the strange structure that $\mathbb{Z}_2\times\mathbb{Z}_2$ has?
- Would this problem be solvable by using the First Isomorphism Theorem?
- I am pretty sure about the existence of $\varphi$, but how would I prove that this is indeed a homomorphism?
- How do I prove that $\ker \varphi = H$?
Sorry for my bad English, and thanks in advance. Any help would be greatly appreciated.
EDIT
I interpreted $\mathbb{Z}_2\times\mathbb{Z}_2$ as the Klein-$4$ group, the 'famous' normal subgroup of alternating group $A_4$. That is why I wrote $\varphi(a), \varphi(b)$ as a permutation. I hope that is not too confusing.