My attempt : Since N is a finite subgroup then S should also be finite . An element of N will be of the form $s^r$ for some s belonging to S . If N normalizes in G for some g in G then $gs^rg^{-1}= s1^m $for s and s1 in S .$ (gsg^{-1})^{r}= s1^m . gsg^{-1} = s1^{m-r}$which shows that $gsg^{-1}$ is in N . If $gSg^{-1}= s1^r$ for then$ (gSg^{-1})^{k} =( s1^r)^{k}$ where k is a natural number which shows that that the element g normalizes for all element of N . I am a bit confused whether I am correct or not ..
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1"An element of $N$ will be of the form $s^r$ for some $s$ belonging to $S$ ": not necessarily so. – Angina Seng Dec 07 '19 at 07:28
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But N is generated by S so will an element not be of the form$ s^r$ ?isn't N a cyclic group ? – Antimony Dec 07 '19 at 07:33
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1No and no${}{}$. – Angina Seng Dec 07 '19 at 07:35
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1$N= \Bigl{s_1^{\epsilon_1}\cdots s_m^{\epsilon_m}\mid m\geq 0,\ s_i\in S,\ \epsilon_i\in{1,-1}\Bigr}$. Please see this post for more details. – cqfd Dec 07 '19 at 07:38
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So if an element g normalizes in G for some g then $gs1^{+-1}s2^{+-1}...sn^{+-1}=a1^{+-1}a2^{+-1}..am^{+-1}g$ which then implies that $gs1^{+-1}..sn^{+-1}g^{-1} = a1^{+-1}a2^{+-1}..an^{+-1} $ . I am getting something like this where the elements s1,..sn and s1,. am are all in S . Does this imply that $gSg^{ -1} $belong to N? – Antimony Dec 07 '19 at 13:22