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$H$ is the set of permutations where $H$ = {$ID_{S_n}$,(12),(34),(12)(34),(13)(24),(14)(23),(1432),(1234)}.

Is $H$ a subgroup of $S_4$?

Is there a simpler way to do this than checking for combinations that may not be closed under the operation? (composition is the operation in permutation groups, right?)

I find that (1432)(12) = (1)(243) = (243) $\notin H$. Is that enough to prove it's not a subgroup or am I testing the elements incorrectly?

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  • $\begingroup$ Well, for starters, $H$ doesn't contain an identity element.... $\endgroup$ – user61527 Nov 19 '13 at 1:15
  • $\begingroup$ oops. I forgot to include it. let me edit that. That was a typo. $\endgroup$ – Tyler Murphy Nov 19 '13 at 1:16
  • $\begingroup$ Well, you should have that $(1432)(12)=(243),$ but apart from that, you did fine. $\endgroup$ – Cameron Buie Nov 19 '13 at 1:18
  • $\begingroup$ Ugh. I'm having a serious typo problem between my whiteboard and keyboard. I did have that. Thanks for the catch. $\endgroup$ – Tyler Murphy Nov 19 '13 at 1:22
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Showing it isn't closed under the group operation with an example, such as you did, correctly shows it is not a subgroup.

Alternatively, you can note that $H$ contains $(1,2)$ and $(1,2,3,4)$, and therefore generates $S_4$. In particular, if it were a subgroup, then $H=S_4$. But it clearly has less than $|S_4|=24$ elements.

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  • $\begingroup$ It appears that the lack of identity was a typo, not intentional. But the second point is correct. $\endgroup$ – user61527 Nov 19 '13 at 1:17
  • $\begingroup$ Is it true then that for $S_n$, if $H$ has a transposition and an $n$-cycle then it generates $S_n$? I hadn't learned that and was wondering if your statement can be generalized. Or is it specific to $S_4$? $\endgroup$ – Tyler Murphy Nov 19 '13 at 1:21
  • $\begingroup$ @TylerMurphy Not quite in general. See the answer here. I adjusted my answer to be less misleading (it's easy to mistake the special case at the end with the general case the answer starts with). Most of the time any transposition and n-cycle will work, but not always. Actually, it turns out that most of the time any two randomly chosen elements will generate all of $S_n$, but this requires advanced techniques. $\endgroup$ – zibadawa timmy Nov 19 '13 at 3:52

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