This is a homework problem and I'm having trouble just getting a basic understanding. I understand that $S_6$ is a symmetric group of degree $6$. I'm not sure how to start looking for elements.
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At the very least you know $(1)$ will work! – graydad Sep 07 '14 at 22:06
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So, I would look at elements as: 1, 2, 3, 4, 5, 6, 12, 13, 14, 15, 16, 23, 24, 25, 26, 34, 35, 36, 45, 46, 56, 123, 124, 12*5, and so on? Then once I know all of my elements, I can determine which ones will commute with (12)(34)(56)? – aneorddot Sep 07 '14 at 22:32
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Not quite.. My abstract algebra is a little rusty but I'll do my best to explain. Regardless of the permutation group, (1) is always the identity permutation. The way it was explained to me is that (1) takes any element in a permutation group and does nothing to it. That means given $n$ and $\sigma \in S_{n}$, it is always true that $$(1)\sigma=\sigma =\sigma (1)$$ so (1) commutes. That's about all I can contribute to your question unfortunately. It looks like some other users have provided helpful hints below though! – graydad Sep 07 '14 at 23:00
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The given element $g = (12)(34)(56) = \sigma_1 \sigma_2 \sigma_3$ with all $[\sigma_i, \sigma_j] = 1$. Hence anything in $\langle \sigma_1, \sigma_2, \sigma_3\rangle$ commutes with $g$.
anomaly
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The permutation $\sigma= \left( \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6\\ a_1 & a_2 & a_3 & a_4 & a_5& a_6 \end{array} \right) $ commutes with the permutation $(1,2)(3,4)(5,6)$ if and only if the permutation $(a_1 a_2)(a_3a_4)(a_5a_6)$ is the same as the permutation $(12)(34)(56)$, that is, if and only if $(a_1 a_2)(a_3a_4)(a_5a_6)$ is a rearrangement of $(12)(34)(56)$. There are $2^3 \times 3! = 48$ such permutations. For instance, $\ \sigma= \left( \begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6\\ 4 & 3 & 1 & 2 & 5 & 6 \end{array} \right) $ commutes with $(12)(34)(56)$ since $(43)(12)(56)= (12)(34)(56)$.
The general answer is not too far.
orangeskid
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