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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 $4$.

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    $\begingroup$ yup, looks good to me $\endgroup$
    – hunter
    Commented Dec 15, 2020 at 6:17

2 Answers 2

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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.

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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 $(14)(23)$.

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