# Find the Number of Group Homomorphisms from $\mathbb{Z}/10\mathbb{Z}$ to $A_4$.

Is this reasoning correct?

$$\mathbb{Z}/10\mathbb{Z}$$ is isomorphic to $$\mathbb{Z}_{10}$$ which is generated by $$1$$, hence we may look at the number of homomorphisms between $$\mathbb{Z}_{10}$$ and $$A_4$$ for the answer to the original question. A homomorphism $$\phi$$ must map the element $$1$$ of $$\mathbb{Z}_{10}$$ to an element of $$A_4$$ such that its order divides both $$10$$ (by properties of homomorphisms) and $$12$$ (by Lagrange's Theorem). Thus, the only possible candidates for the mapping of $$1$$ must have order $$1$$ or $$2$$. This in turn yields $$4$$ possible elements in $$A_4$$, namely, $$()$$, $$(12)(34)$$, $$(13)(24)$$, and $$(14)(23)$$. All of these mappings preserve the group operation, thus the answer to the original question is $$4$$.

• yup, looks good to me Commented Dec 15, 2020 at 6:17

Yes, your reasoning is correct. Yet another way to tackle the problem is by the fundamental homomorphism theorem. Here we are looking for the (number of) possible subgroups of $$A_4$$ isomorphic to quotients of $$\Bbb Z_{10}$$; the (normal) subgroups of $$\Bbb Z_{10}$$ are: $$\{0\}$$, $$\{0,5\}\cong \Bbb Z_2$$, $$H:=\{0,2,4,6,8\}\cong \Bbb Z_5$$ and $$\Bbb Z_{10}$$; for order reasons (Lagrange theorem), the only options are $$\Bbb Z_{10}/\Bbb Z_{10}\cong\Bbb Z_1$$ and $$\Bbb Z_{10}/H\cong\Bbb Z_2$$. Therefore, if we call $$\phi$$ such a homomorphism, we can have either $$\phi(\Bbb Z_{10})=\{()\}$$ (the trivial case) or any among $$\phi(\Bbb Z_{10})=\{(),(12)(34)\}$$, $$\phi(\Bbb Z_{10})=\{(),(13)(24)\}$$, $$\phi(\Bbb Z_{10})=\{(),(14)(23)\}$$, so $$4$$ altogether.
Your argument is correct. A slightly different way to see it is to note that choosing an element $$\sigma\in A_4$$ defines uniquely a homomorphism $$f_\sigma\colon\mathbb{Z}\to A_4$$ by $$f_\sigma(n)=\sigma^n$$ and this induces a homomorphism $$\mathbb{Z}/10\mathbb{Z}\to A_4$$ if and only if $$10\mathbb{Z}\subseteq\ker f_\sigma$$ Conversely, a homomorphism $$\mathbb{Z}/10\mathbb{Z}\to A_4$$ induces a homomorphism $$\mathbb{Z}\to A_4$$ by composing with the canonical projection.
The kernel of $$f_\sigma$$ is easy to compute: it is $$k\mathbb{Z}$$ where $$k$$ is the order of $$\sigma$$, because the image of $$f_\sigma$$ is precisely $$\langle\sigma\rangle$$.
Now we know that $$k\mathbb{Z}\supseteq 10\mathbb{Z}$$ if and only if $$k\mid 10$$ and so the order of $$\sigma$$ can only be $$1$$ or $$2$$. This makes for just four choices of $$\sigma$$, namely $$()$$, $$(12)(34)$$, $$(13)(24)$$ and $$(14)(23)$$.