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$\begingroup$

I am looking for examples of socle and normal-subgroup relations.

If $G=S_{4}$, the normal subgroups are $A_{4}$ and $V_{4}$, thus the socle of $G$ is the klein-four group, and for $A_{4} \cap Soc(G) = Soc(A_{4}) = V_{4}$.

Now I am looking for some easy example where this does not work. Where we have no normal subgroup in $G$, s.t. the cut with the $Soc(G)$ equals the socle of this normal subgroup.

Any hints where I can look for such a group?

Thanks and best :) Kathrin

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    I think you are out of luck, because ${\rm Soc}({\rm Soc}(G))={\rm Soc}(G)$.2012-11-21
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    I mean like for a normal subgroup $N \unlhd G$ s.t. $N \cap Soc(G) \not= Soc(N)$. Therefore I want to find a group and normal subgroup that fulfill this. Then there is no $Soc(Soc(G))$ right?2012-11-21
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    Let $G$ be dihedral of order 8, which has socle of order 2, and $N$ a normal subgroup isomorphic to $C_2^2$, which has socle of order 4.2012-11-21
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    First thanks :)
    $G = \{e, T, T^{2}, T^{3}, S, ST, ST^{2}, ST^{3}\}$ where $T$ is a turn of $\frac{\pi}{2}$ and $S$ the mirrow.

    Then I have the normal subgroups: $e, , , , $. The minimal normal subgroup is right? as this is contained in all the other normal subgroups. So $Soc(G) = = \{e, T^{2}\}$. But now for all normal subgroups the intersection is $$. But isn't also the Socle of all these normal subgroups $$?
    2012-11-21
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    No, the socle of the normal subgroup $\langle S,T^2 \rangle$ is equal to itself (order 4), and the same applies to the normal subgroup $\langle ST,T^2 \rangle$.2012-11-21
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    Ah ok, so for $N = $ for example, the minimal normal subgroups are $$ and $$ s.t $Soc(N) = \{e, S, T^{2}, ST^{2}\}$? Thanks a lot for all the help :)2012-11-21

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