# Show that the set ${e, \sigma}$ is a subgroup of $S_3$, where e is the identity element of $S_3$.

Consider the symmetric group $S_3$ and the permutation $\sigma = (1, 2)$.

Show that the set ${e, \sigma}$ is a subgroup of $S_3$, where e is the identity element of $S_3$.

I've begun to attempt this by using the subgroup test but i'm really not sure how to apply it to this question.

"The Sub Group Test:

Let H be a subset of a group G. Then H is a subgroup of G if and only if both of the following are satisfied.

i) H is nonempty.

ii) $ab^{−1} ∈ H$, for all $a, b ∈ H$."

• This "subgroup test" that you wanted to apply, what is that exactly? – Arthur Dec 6 '17 at 14:38
• What parts of the subgroup test have you checked? Where are you stuck? Please edit the question to show your work. – Ethan Bolker Dec 6 '17 at 14:38
• Write out the Cayley table and stare at it. – Randall Dec 6 '17 at 14:46
• @Arthur I have included the Sub group test in the question now. – Ben Jones Dec 6 '17 at 14:52
• Now show us how you tried to check (i) and (ii) in your problem. – Ethan Bolker Dec 6 '17 at 14:54

This is small enough that we can simply consider all the possible products, choosing each of $a$ and $b$ to be either $e$ or $(12)$:
1. $a = e, b = e$ gives $ee^{-1} = e$, which is in the subset
2. $a = e, b = (12)$ gives $e(12)^{-1} = (12)$, which is in the subset
3. $a = (12), b = e$ gives $(12)e^{-1} = (12)$, which is in the subset
4. $a = (12), b = (12)$ gives $(12)(12)^{-1} = e$, which is in the subset