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?

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.
