# Determine the number of homomorphisms between $\Bbb{Z}_{10} \times \Bbb{Z}_{25}$ and $S_4$

Determine the number of homomorphisms between $\Bbb{Z}_{10} \times \Bbb{Z}_{25}$ and $S_4$.

Here, $\Bbb{Z}_n$ is the integers from $0$ to $n-1$ with addition modulo $n$. $S_4$ is the symmetric group (permutations of $\{1,2,3,4\}$).

I think the number of homomorphisms between these groups is $9$, can somebody verify this for me?

EDIT: My reasoning is that $\langle(1,0),(0,1)\rangle$ generates the above product group and the order of these elements are 10 and 25 respectively. So, the order of the elements which these elements are mapped to in $S_4$ must divide 10 and 25 respectively.

Ie.

$o((1,0)) \in \{1,2,5,10\}$

$o((0,1)) \in \{1,5,25\}$

Now considering $S_4$, I think there is one element of order $1$ (identity) and $9$ elements of order $2$ (cycle types 2 + 1 + 1 [6] and 2 + 2 [3], right?) There aren't any elements of order of the other divisors. So, the possibilities are that the first element above can map to $9$ elements and the second can only map to the identity, so, $9$ homomorphisms in total?

Anyone spot a mistake in my reasoning?

Thanks!

• There are more elements in $S_4$ of order $2$ than just the transpositions, for example $(1, 2, 3, 4) \mapsto (2, 1, 4, 3)$. – Anon May 12 '16 at 21:20
• @McFry EDIT: Apologies, I thought you were saying that there are other elements with an order other than 1 or 2, I see, the permutation you have above actually decomposes to two transpositions and has order 2. – Helen Byrne May 12 '16 at 21:45
• In that case, are the elements of order 2 the permutations with only a single transposition (cycle type 2 + 1 + 1) and then permutations of cycle type 2 + 2 (two transpositions)? So, is it 9 elements of order 2? – Helen Byrne May 12 '16 at 21:51

Your reasoning is correct, except that you've miscounted the number of elements of $S_4$ of order $2$. There should be $6$ single transpositions and $3$ products of disjoint transpositions.