I know that, using the first isomorphism theorem, I have to define a suprajective homomorphism between $\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$ such that $\ker f = \{([0]_5,[0]_3), \;([0]_5,[1]_3), \;([0]_5,[2]_3)\}$ but I haven’t find it because every correspondence I come up with is not even a function. Any hint?
2026-03-26 14:42:56.1774536176
Is $(\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z})/\left<([0]_5,[2]_3)\right> \cong \mathbb{Z}/2\mathbb{Z} $?
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Hint: Note that we can find an isomorphism between $(\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z})/(\;([0]_5,[2]_3)\;)$ and $\Bbb Z/ 5\Bbb Z$. So, the statement that you are trying to show is equivalent to saying that $\Bbb Z/5 \Bbb Z \cong \Bbb Z / 2\Bbb Z$. What is an easy way to see that these two rings (or groups) are not isomorphic?