Calculating elements of the $\mathbb{Z}^2/\!\ker\varphi$ group

Question

If I know that $\varphi: \mathbb{Z}^2 \to S_{15}$ is the homomorphism defined as follows: \begin{align} \varphi(1,0)&=(2,7,3)(3,11,5)(12,13)\\ \varphi(0,1)&=(14,15)(1,4,6,8,9,10) \end{align} I was asked to calculate how many elements $\mathbb{Z}^2/\!\ker\varphi$ has.

Is it true to say that from the first homomorphism theorem we get $\mathbb{Z}^2/\!\ker\varphi \cong S_{15}$. So from $|S_{15}|=15!$ we can understand that $\mathbb{Z}^2/\!\ker\varphi$ has $15!$ elements. For some reason, it is written in the textbook that is has only $60$ elements. Why is that?

Answer 1

To apply the homomorphism theorem the map should be an epimorphism.

In the other hand to count how many elements in the image are, you just got to see is the orders of $\varphi(1,0)=(2,7,3,11,5)(12,13)$ and $\varphi(0,1)$ which happen to be ten and six respectively. Then $|{\rm im}\varphi|$ is $60$.

Update:

The next calculations might help you to understand \begin{eqnarray*} \varphi(2,0)&=&((2,7,3,11,5)(12,13))^2,\\ &=&(2,3,5,7,11).\\ &&\\ \varphi(3,0)&=&((2,7,3,11,5)(12,13))^3,\\ &=&(2,11,7,5,3)(12,13).\\ &&\\ \varphi(4,0)&=&(2,5,11,3,7).\\ &&\\ \varphi(5,0)&=&(12,13).\\ &&\\ &...&\\ &&\\ \varphi(10,0)&=&e. \end{eqnarray*}

Answer 2

Set, for simplicity, $\sigma=\varphi(1,0)$ and $\tau=\varphi(0,1)$. Then $$ \sigma=(2,7,3,11,5)(12,13) $$ and so $\sigma$ and $\tau$ are disjoint, so $\sigma\tau=\tau\sigma$ and the homomorphism is indeed well defined.

The image of $\varphi$ is an abelian subgroup of $S_{15}$, so it certainly isn't the whole $S_{15}$.

Set, for simplicity, $\sigma=\varphi(1,0)$ and $\tau=\varphi(0,1)$. Then $$ \varphi(m,n)=\sigma^m\tau^n $$ and so the image of $\varphi$ is the subgroup generated by $\sigma$ and $\tau$. Now you should be able to finish, using the fact that there is an obvious surjective homomorphism $$ \langle\sigma\rangle\times\langle\tau\rangle\to \langle\sigma\rangle\langle\tau\rangle=\operatorname{im}\varphi $$ What's the kernel of this homomorphism?