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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 ${\rm Soc}({\rm Soc}(G))={\rm Soc}(G)$. – Derek Holt Nov 21 '12 at 15:42
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? – Kathrin Nov 21 '12 at 15:48
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. – Derek Holt Nov 21 '12 at 16:41
First thanks :) <br/> $G = \{e, T, T^{2}, T^{3}, S, ST, ST^{2}, ST^{3}\}$ where $T$ is a turn of $\frac{\pi}{2}$ and $S$ the mirrow. <br/> <br/> Then I have the normal subgroups: $e, <T>, <T^{2}>, <S,T^{2}>, <ST, T^{2}>$. The minimal normal subgroup is <T^{2}> right? as this is contained in all the other normal subgroups. So $Soc(G) = <T^{2}> = \{e, T^{2}\}$. But now for all normal subgroups the intersection is $<T^{2}>$. But isn't also the Socle of all these normal subgroups $<T^{2}>$? – Kathrin Nov 21 '12 at 17:17
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 $\langle ST,T^2 \rangle$. – Derek Holt Nov 21 '12 at 17:55

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