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$G=\mathbb{Z}_2\times\mathbb{Z}_3$ 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.

  • 0
    I find the statement of Condition 4 completely ambiguous! It is true with one interpretation and false with the other.2012-07-13

4 Answers 4

3

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 $6$?

2

Hint: consider permutations of $\{1,2,3,4,5\}$ that leave a certain set invariant.

2

HINT: Consider the permutation $(12)(345)$. (Compare it with $\Bbb Z_2\times\Bbb Z_3$.)

1

Your group is isomorphic to $Z_6$. I think $S_5$ has an element of order $6$, i.e. $(123)(45)$.