Finding a finite group that is a quotient of $\pi_1 T$ but nota quotient of $\pi_1 K$ Let $T$ be the torus and $K$ the Klein bottle. The fundamental groups of these spaces are
$$
\pi_1 T \cong\langle a,b \mid aba^{-1}b^{-1}\rangle,\qquad \pi_1 K \cong\langle a,b \mid aba^{-1}b\rangle
$$
I am trying to find a finite group that is a quotient of $\pi_1 T$ but not of $\pi_1 K$.
Any hints on how I should go about proving this? For instance, is there a way to show that $\mathbb{Z}_n\times \mathbb{Z}_m$ is not a quotient group of $\pi K$.
My attempt following @JoshuaP.Swanson 's comment:
I will use the fact that $\pi_1T\cong \mathbb{Z}\times\mathbb{Z}$ and $\pi_1K = \mathbb{Z}\rtimes\mathbb{Z}$.
Suppose by way of contradiction that $[\mathbb{Z}\rtimes\mathbb{Z}]/N \cong \mathbb{Z}_m\times\mathbb{Z}_n$ is a quotient group of $\mathbb{Z}\rtimes\mathbb{Z}$. Then $N$ is a normal subgroup of $\mathbb{Z}\rtimes\mathbb{Z}$ and since $[\mathbb{Z}\rtimes\mathbb{Z}]/N$ is abelian it must contain all elements of $\mathbb{Z}\rtimes\mathbb{Z}$ of the form $(a,b)(a',b')(a,b)^{-1}(a',b')^{-1}$ where $(a,b), (a',b')\in \mathbb{Z}\rtimes\mathbb{Z}$. We can compute these elements explcitly (am I doing this right?):
$$
(a,b)(a',b')(a,b)^{-1}(a',b')^{-1} = \begin{cases}
(0,0) &\text{if }b,b'\text{ are even}\\
(2(a+a') &\text{if }b,b'\text{ are odd}\\
(2a,0) &\text{if }b\text{ is even even and }b'\text{ is odd}\\
(2a',0) &\text{if }b\text{ is even odd and }b'\text{ is even}.
\end{cases}
$$
In particular, $N\supseteq 2\mathbb{Z}\times \{0\}$. Does this lead to a contradiction?
 A: The fundamental group of the Klein bottle is the semidirect product of two $\mathbb{Z}$s where the first acts on the second by the involution which sends $1$ to $-1.$  This would seem to indicate that the dihedral groups are quotients of $\pi_1(K),$ but the direct products $\mathbb{Z}/p \mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ are not.
A: I'll expand on my hint from the comments. I recommend you read only as much as you need until you're able to solve it.
You seem to be roughly near the right track, though I think you could benefit from some more conceptual understanding of the abstraction. For instance, the relation $aba^{-1}b^{-1}$ is really saying you're imposing the relation $aba^{-1}b^{-1} = 1$, i.e. $ab = ba$, and nothing else, which is why you get $\pi_1 T \cong \mathbb{Z} \times \mathbb{Z}$. Clearly this is abelian already, so its abelianization is itself. Any quotient is hence going to be abelian.
Likewise the relation $aba^{-1}b$ is really saying $aba^{-1} = b^{-1}$, which is where the semidirect product is coming from. But now suppose we were looking at this relation in an abelian quotient. We'd have $\bar{a}\bar{b}\bar{a}^{-1} = \bar{b}^{-1}$, but since the quotient is abelian we'd also have $\bar{a}\bar{b} = \bar{b}\bar{a}$, i.e. $\bar{a}\bar{b}\bar{a}^{-1} = \bar{b}$, hence $\bar{b} = \bar{b}^{-1}$. This suggests we consider $\mathbb{Z} \times \mathbb{Z}/2$ as a potential abelianization of $\pi_1 K$.
To check that, we can first let $\pi_1 K \to \mathbb{Z} \times \mathbb{Z}/2$ by $a \mapsto (1, 0)$, $b \mapsto (0, 1)$. Note that the relation $aba^{-1} = b^{-1}$ becomes $(1,0) + (0, 1) + (-1, 0) = (0, 1)$, so the homomorphism is well-defined. Obviously the homomorphism is surjective.
Moreover, the calculation we just did above says exactly that if we have some abelian quotient $Q$ with natural projection map $\pi_1 K \to Q$ given by $a \mapsto \overline{a}$, $b \mapsto \overline{b}$, then $\overline{b} = \overline{b}^{-1}$. You should check that this says the homomorphism $\mathbb{Z} \times \mathbb{Z}/2 \to Q$ given by $(1, 0) \mapsto \overline{a}$, $(0, 1) \mapsto \overline{b}$ is well-defined. It is obviously surjective. By the first isomorphism theorem, that says $Q$ is a quotient of $\mathbb{Z} \times \mathbb{Z}/2$. This is a manifestation of the mantra that the abelianization is the largest abelian quotient as well as an explicit version of the universal property of the abelianization. Note that I avoided explicitly computing the commutator subgroup.
So, now you just have to find a finite quotient of $\mathbb{Z} \times \mathbb{Z}$ that is not a quotient of $\mathbb{Z} \times \mathbb{Z}/2$.
