
#1
November 23rd, 2016,
12:55
Consider the subset of $S_4$ defined by
$$K_4=\{(1)(2)(3)(4),(12)(34),(13)(24),(14)(23)\}$$
Show that for all $f \in K_4$ and all $h \in S_4$, we have $h^{1}fh \in K_4$
I showed all the possible cycle shapes of h and am trying to show that $h^{1}fh$ must always have cycle shape $(2,2)$, excluding the case of identity permutation.
Just don't know where to go from here
Last edited by Euge; November 23rd, 2016 at 14:09.
Reason:
improve formatting

November 23rd, 2016 12:55
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#2
November 23rd, 2016,
14:06
Hi, Confusedalways! Welcome.
You're on the right track. Consider the following fact. If $\sigma = (a_1 \cdots a_r)$ is a cycle in $S_n$ and $\tau\in S_n$, then $\tau \sigma \tau^{1} = (\tau(a_1)\cdots \tau(a_r))$. Take for instance $(12)(34)$. For all $h\in S_4$, $$h(12)(34)h^{1} = h(12)h^{1}h(34)h^{1} = (h(1)\;h(2))\,(h(3)\;h(4))$$
so $h(12)(34)h^{1}$ has cycle structure $(2,2)$. The same argument applies to the others.