Permutation Representations of Elements and Cycle Length of Conjugates

Question

Good morning. I am looking at the permutation representations of the group action by conjugation: Let $g,a\in G$. Define a left action on G by

$$g\cdot a=gag^{-1}$$

I'm specifically looking at $D_8$. I know that the permutation representation of the elements of $D_8$ are

$$\sigma_e=(1), \sigma_r=(1234), \sigma_{r^2}=(13)(24), \sigma_{r^3}=(1432) $$ $$\sigma_s=(24),\sigma_{sr}=(14)(23),\sigma_{sr^2}=(13),\sigma_{sr^3}=(12)(34)$$

Both $e_{D_8}, r^2\in Z(D_8)$. As proposed in Dummit and Foote,

Two elements of $S_n$ are conjugate in $S_n$ iff they have the same cycle type. The number of conjugacy classes of $S_n$ equals the number of partitions of $n$

But $r^2$ has the same cycle type as $sr$ and $sr^3$. Why are they not conjugate? Is it simply because $r^2\in Z(D_8)$? Does the second part of the proposition verify that? Partitions of $4$ are

$$1+1+1+1, 1+1+2, 1+3, 4, 2+2$$

Answer 1

If two elements of a subgroup are conjugate to each other in the larger group, this implies nothing about conjugacy in the subgroup.

For example, you could represent the Klein $4$-group as $$\{(1),(12),(34),(12)(34)\}$$ Being of the same cycle type, $(12)$ and $(34)$ are conjugate in the symmetric group. However this subgroup is abelian so they are obviously not conjugate in the subgroup.

Answer 2

As noted by @EuYu, the Theorem applies in $S_4$, not $D_8$.

$r^2$, $sr$ and $sr^3$ are conjugate in $S_4$, but not in $D_8$. You can use the Theorem however to rule conjugacies out - if $x$ and $y$ are conjugate in $G\le S_n$ then they're conjugate in $S_n$ so have the same cycle type. In this example, $r$ and $sr^2$ for example cannot be conjugate.

Also, as you note, $r^2\in Z(D_8)$, so you know it is the only element of it's conjugacy class. This means that in order to work out all conjugacy classes you just need to check whether the following pairs are conjugate:

$(r,r^3),(s,sr^3),(sr,sr^3)$