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.

Thanks in advance.

"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 $a, b ∈ H$."

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

I guess you can do the first criterion of your subgroup test easily enough. So I'll look at the second one.

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

This confirms criterion ii) of your subgroup test, so the subset is indeed a subgroup.


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