# Is $\Bbb{Z}/4\Bbb{Z}\times S_3$ isomorphic to $S_4$?

Is $\Bbb{Z}/4\Bbb{Z}\times S_3$ isomorphic to $S_4$?

They both have the same number of elements (i.e. have order $24$), and they both are non-cyclic and non-albelian, but I am unsure what other properties to check unless I start counting the order of every single element in both groups (i.e. a total of $48$ calculations). Is there a quicker way to prove or disprove whether they are isomorphic?

• Just by looking at cycle types, it's not hard to see every element of $S_4$ has order at most $4$. Can you find an element of order greater than $4$ in $\mathbb{Z}/4\mathbb{Z}\times S_3$? – Ben West Nov 2 '17 at 7:37
• Thank you would you be able to explain where "every element of S4 order at most 4" comes from? I know the individual order of an element of a group must divide the total order of the group. Which means the elements of S4 have order 1,2,3,4,6,8. And its not mentioned on wikipedia that Sn permutations must have order at most n so where is this derived from? Thank you :) – E.J Humphrey Nov 2 '17 at 9:05
• Sorry I should make my query above more clear. For example it is easy to see that $$(1 2 3 4)^4 = (1)$$ but where are the theorems to prove this without testing every single element in S4 to make sure there is no permutation with order greater than 4? – E.J Humphrey Nov 2 '17 at 9:16
• The key facts are that a cycle of length $n$ has order $n$, and the order of a product of disjoint cycles is the least common multiple of the lengths of the cycles. The only cycle types in $S_4$ are $[1],[2],[2,2],[3],[4]$ (I've written them with [ ] to not be confused with actual permutations). To their respective orders are $1,2,2,3,4$. Be aware in $S_5$ for example, an element of cycle type $[2,3]$ has order $\operatorname{lcm}(2,3)=6$, so in general $S_n$ may have elements of order greater than $n$. – Ben West Nov 2 '17 at 18:18

The center of $S_4$ is trivial, whereas the center of $\Bbb Z/4\Bbb Z \times S_3$ is not.
$\Bbb Z/4\Bbb Z\times S_3$ has an element of order $12$; $S_4$ hasn't.
$\mathbb{Z}/4\mathbb{Z}\times S_3$ has a unique subgroup of order $3$ whereas $S_4$ has four of them.
$\mathbb{Z}/4\mathbb{Z}\times S_3$ has an abelian Sylow-2 subgroup whereas the Sylow-2 subgroup of $S_4$ is non-abelian (in fact, it's a $D_8$).
$\mathbb{Z}/4\mathbb{Z} \times S_3$ has a cyclic normal subgroup of order $4$, whereas any cyclic subgroup of order $4$ of $S_4$ is generated by a $4$-cycle, say $(a_1 a_2 a_3 a_4)$, but if the subgroup has to be normal then it must also contain $(a_2 a_1 a_3 a_4)$, a contradiction.