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Let $G=\langle a_{1}\rangle\times\langle a_{2}\rangle$ such that $|a_{i}|=2^{k_{i}}$ and $k_{1}>k_{2}$ and $H$ be a subgroup of $G$ that there exists an automorphism of $G$ such that fix only the elements of $H$. How do we can find all subgroups of $G$ with this property?

Note: Any subgroup $H$ with the above property must contain the element $a_{1}^{2^{k_{1}-1}}$.

please guide me. Thanks

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  • $\begingroup$ I suppose this is related to your MSE question math.stackexchange.com/questions/851243/… . Is this follow-up purely from interest, or from a similar exercise? $\endgroup$ Jul 10, 2014 at 7:56
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    $\begingroup$ I think that when $H \le \langle a_1 \rangle$ it is possible if and only if $|H| \ge 2^{k_1-k_2}$. $\endgroup$
    – Derek Holt
    Jul 10, 2014 at 8:31
  • $\begingroup$ @DerekHolt: Yes, very nice. All examples that i investigated, confirm your comment. $\endgroup$
    – elham
    Jul 10, 2014 at 9:33

1 Answer 1

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As your example below shows, we cannot assume in all cases that $H \le \langle a_1 \rangle$ as I asserted previously, but I'll explain how to do that case anyway.

To avoid all the notation, I will consider two specific cases, one where it is not possible and one where it is possible. That should be enough to enable you to solve it in general. I am going to call the generators $a,b$ rather than $a_1,a_2$.

First suppose that $|a|=32$, $|b|=4$, and $|H|=4$, so $H = \langle a^8 \rangle$. Suppose $\phi \in {\rm Aut}(G)$ with ${\rm Fix}(\phi) = H$. Since $\phi(b^2) \ne b^2$ or $a^{16}$, we must have $\phi(b^2) = a^{16}b^2$. Also, since $\phi(a^8) = a^8$, we must have $\phi(a^4)^2 = a^8$, and so $\phi(a^4) = a^4, a^{20}, a^4b^2$ or $a^{20}b^2$. The first is not possible, since $a^4 \not\in {\rm Fix}(\phi)$. If $\phi(a^4) = a^{20}$, then $\phi(a^4b^2) = a^4b^2$, which is not possible because $a^4b^2 \not\in {\rm Fix}(\phi)$. The other two possibilities are ruled out by the fact that $a^4$ is a $4$-th power, but neither $a^4b^2$ nor $a^{20}b^2$ is a $4$-th power. So there i no such $\phi$ in this case.

Now suppose that $|a|=32$, $|b|=4$, and $|H|=8$, so $H = \langle a^4 \rangle$. Then we can define $\phi$ with ${\rm Fix}(\phi) = H$ by $\phi(a) = a^{25}b$, $\phi(b)=a^8b^3$.

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  • $\begingroup$ Thank you. Only one question. for any subgroup $H$ we can we can choose the generators $a_{1}, a_{2}$ of $G$ such that $H=H_{1}\times H_{2}$ with $H_{i}\leq ⟨a_{i}⟩$. For my better understand of this, please explain this special example: Let $G=\langle a\rangle\times\langle b\rangle$ $|a|=16$ , $|b|=2$ and $H=\langle a^4b\rangle$.Then how do can choose the generators $a_{1}$ and $a_{2}$ for $G$ such that $G=\langle a_{1}\rangle\times\langle a_{2}\rangle$ and $H=H_{1}\times H_{2}$ with $H_{i}\leq ⟨a_{i}⟩$. please guide me. $\endgroup$
    – elham
    Jul 10, 2014 at 15:28
  • $\begingroup$ No you can't do it that case! Sorry,that was a mistake. $\endgroup$
    – Derek Holt
    Jul 10, 2014 at 17:08

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