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To prove $r^2 \in N_G(\langle s,r^4\rangle)$, Why we don't need to consider $(sr^4)^2, (sr^4)^3,\ldots,(sr^4)^n$ and $s^nr^{4m}$ where $n\neq m$ are integers. That is, the multiple combinations of $s$ and $r^4$.

That is, why don't we need to consider the result $r^2(sr^4)^2r^{-2}$, $r^2(sr^4)^3r^{-2}$, $r^2(sr^4)^4r^{-2},\ldots$ and $r^2(s^nr^{4m})r^{-2}$?

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    $\begingroup$ I changed $<s,r^4>$ to $\langle s,r^4\rangle$. That is standard usage. $\endgroup$ Feb 28 '13 at 4:54
  • $\begingroup$ Also, it's preferable to get screenshots (or type in the content yourself) so that when link rot inevitably occurs, stuff here is still intelligible. $\endgroup$ Feb 28 '13 at 4:56
  • $\begingroup$ Can anyone answer? $\endgroup$
    – Ian
    Feb 28 '13 at 5:24
  • $\begingroup$ @Ian: Can you bring me the definition of $D_{16}$ here? $\endgroup$
    – Mikasa
    Feb 28 '13 at 5:27
  • $\begingroup$ @Babak S.:It's the diherdal group of order 16. $D_{16}$={$r,s|r^8=s^2=1$, $rs=sr^{-1}$} $\endgroup$
    – Ian
    Feb 28 '13 at 5:32

Note that if $H$ is a subgroup and $gag^{-1}, gbg^{-1} \in H$ then automatically we have $gabg^{-1} \in H$ because $gabg^{-1} = (gag^{-1})(gbg^{-1})$. Thus once you have tested that conjugation does not take $sr^4$ outside of your subgroup it is then automatic that for any $n$ it does not take $(sr^4)^n$ out of your subgroup.

In general to test that $gHg^{-1} \subseteq H$ is suffices to test that $gag^{-1} \in H$ for each $a$ in a set of generators of $H$.


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