Given a finite group $G$ and any subgroup $H$. In all the examples I've looked at the intersection of two different conjugate subgroups of $H$ always had the same size. Is this always the case? I know that the intersection $H^{g_1} \cap H^{g_2}$ has the same size as $H \cap H^{g_2g_1^{-1}}$, but that's a bit of how far I got.
1 Answers
$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$There might be a simpler example, still the following one is somewhat natural.
Consider the abelian group $A$ of order $2^{6}$, $$ A = \Span{a_{1}} \times \Span{b_{1}} \times \Span{a_{2}} \times \Span{b_{2}} $$ where $a_{1}, a_{2}$ have order $4$, and $b_{1}, b_{2}$ have order $2$. Note that $c_{1} = a_{1} b_{1}$ and $c_{2} = a_{2} b_{2}$ have order $4$.
Consider the automorphism $g$ of $A$ which maps $$ a_{1} \mapsto c_{1}, c_{1} \mapsto a_{1}, a_{2} \mapsto c_{2}, c_{2} \mapsto a_{2}. $$ An alternative description of $g$ is $$ a_{1} \mapsto c_{1}, b_{1} \mapsto b_{1}, a_{2} \mapsto c_{2}, b_{2} \mapsto b_{2}. $$ Consider the automorphism $h$ of $A$ which maps $$ a_{1} \mapsto a_{2}, c_{1} \mapsto c_{2}, a_{2} \mapsto a_{1}, c_{2} \mapsto c_{1}. $$ An alternative description of $h$ is $$ a_{1} \mapsto a_{2}, b_{1} \mapsto b_{2}, a_{2} \mapsto a_{1}, b_{2} \mapsto b_{1}. $$ Then in the natural semidirect product $A \rtimes \Span{g, h}$ the four subgroups $$ \Span{a_{1}}, \Span{c_{1}}, \Span{a_{2}}, \Span{c_{1}} $$ form a conjugacy class, but $$ \Span{a_{1}} \cap \Span{c_{1}} = \Span{a_{1}^{2}} $$ has order two, while $$ \Span{a_{1}} \cap \Span{a_{2}} = \Set{1}. $$
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I already upvoted your answer, once I put the construction into a CAS I'll accept the answer. This can take a while. Moreover the cas construction of the semidirect product gives $((C2 \times ((C4 \times C2) \rtimes C4)) \rtimes C2) \rtimes C2$ as a structure description, making it difficutlt to find the natural embedding of $A$. – Marc Bogaerts Oct 20 '16 at 15:33
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@BogaertsMarc, ok, thanks. Should you need further details - like, why $g$ and $h$ are automorphisms - just let me know. – Andreas Caranti Oct 20 '16 at 15:36
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Yes, I already figured that out, they are well defined of suitable generators of $A$, and together make up the Klein group. I'm just curious what make these subgroups "different" though they are isomorphic. Perhaps they lie in non isomorphic subgroups? – Marc Bogaerts Oct 20 '16 at 15:45