# $G=\mathbb{Z}_2\times\mathbb{Z}_3$ is isomorphic to?

$G=\mathbb{Z}_2\times\mathbb{Z}_3$ is isomorphic to

1. $S_3$

2. A subgroup of $S_4$.

3. A proper subgroup of $S_5$

4. $G$ is not isomorphic to a subgroup of $S_n$ for all $n\ge 3$

What I know is Any finite group is isomorphic to a subgroup of $S_n$ for some suitable $n$(Caley's Theorem), 1 is not true as $G$ is abelian but $S_3$ is not,$4$ violates Caleys Theorem, for 3 I saw that there is an order $6$ element in $G$ namely $(1,1)$ but No element of the subgroup of $S_4$ have order $6$ right? Not sure about 3, Thank you for the help.

• If a group $G$ is isomorphic to a subgroup of $S_n$ for some $n$, then in particular $G$ has to be finite. Therefore NOT ANY group is isomorphic to a subgroup of $S_n$. – Mercy King Jul 11 '12 at 21:54
• @Mercy Unless you interpret $n$ as taking cardinal values as well. – Alex Becker Jul 11 '12 at 21:57
• You might be interested in the following fact: any finite permutation generates a cyclic group of order which is the least common multiple of the disjoint cycles forming the permutation. – tomasz Jul 11 '12 at 21:58
• @Mercy, I am simply pleased and delighted at your comment. – Marso Jul 11 '12 at 22:00
• @tomasz thank you for the comment, sure I will think. – Marso Jul 11 '12 at 22:01

You are correct that $G$ is not isomorphic to $S_3$ but that it is isomorphic t some subgroup of $S_n$ for some $n$ (in fact, any $n\geq 6$ clearly works). Note that $G\cong \mathbb Z_6$, the cyclic group of order $6$. Thus the existence of a subgroup isomorphic to $G$ is equivalent to the existence of an element of order $6$. Any element of $S_4$ is either a $4$-cycle, which has order $4$, a $3$-cycle, which has order $3$, or a product of disjoint $2$-cycles, which have order $2$. Thus $G$ is not isomorphic to a subgroup of $S_4$. Can you think of an element of $S_5$ with order $6$?
Hint: consider permutations of $\{1,2,3,4,5\}$ that leave a certain set invariant.
HINT: Consider the permutation $(12)(345)$. (Compare it with $\Bbb Z_2\times\Bbb Z_3$.)
Your group is isomorphic to $Z_6$. I think $S_5$ has an element of order $6$, i.e. $(123)(45)$.