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I am given that G is a set of elements that commute with (12)(3456) in $S_6$, and asked following questions: i) Show that $G$ is a subgroup of $S_6$.

Attempt( I could just do the first subgroup test, which was trivial): a) So, $e$ commutes with (12)(3456) and thus $e$ $\in$ G, hence G is not empty.

The reason I could not proceed now is because I do not yet know what other elements belong G. Without knowing it, I do not know how to show that it is closed and inverse exists.

Question: How do I know what elements in $S_6$ commute with (12)(3456). Is this question the same as find the elements in $S_6$ that are conjugate with (12)(3456)? I am pretty sure that there is a neat idea to proceed instead of trying every possible combination, but I just do not see it.

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    $\begingroup$ Have you been introduced to centralisers? $\endgroup$ – pureundergrad May 15 '18 at 20:55
  • $\begingroup$ Let $\sigma = (12)(3456)$. Then, you are looking for elements of the form $f \in S_6$ such that $f\sigma = \sigma f$. You want to show that for any such $f$, you have $f^{-1}\sigma = \sigma f^{-1}$. Hint, multiply on the right by $\sigma^{-1}$, then multiply on the left by $f$. Then use the fact that $f$ commutes with $\sigma$. $\endgroup$ – InterstellarProbe May 15 '18 at 20:57
  • $\begingroup$ Yes @pureundersgrad $\endgroup$ – Ufomammut May 15 '18 at 20:58
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    $\begingroup$ There is a more general theorem which says the centraliser is a subgroup, you might want to look it up $\endgroup$ – pureundergrad May 15 '18 at 21:01
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    $\begingroup$ There's nothing special about $(12)(3456)$. And there's nothing special about $S_6$ either. $\endgroup$ – Lord Shark the Unknown May 15 '18 at 21:14
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Let $\sigma = (12)(3456)$. Then, you are looking for elements of the form $ f \in S_6$ such that $f\sigma = \sigma f$. You want to show that for any such $f$, you have $f^{-1}\sigma=\sigma f^{-1}$.

Hint, multiply on the right by $\sigma^{-1}$, then multiply on the left by $f$. Then use the fact that $f$ commutes with $\sigma$.

– InterstellarProbe

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  • $\begingroup$ Not sure how you got the final result. $\endgroup$ – Ufomammut May 15 '18 at 21:11

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