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For a symmetric group $S_n$, we have known clearly the proper nontrivial normal subgroups of it: if n=3 or $\geq 5$, $S_n$ has only one normal subgroup; if n=4, $S_n$ has two normal subgroups; otherwise, $S_n$ has no normal subgroups.

As a generalization of $S_n$ is the signed symmetric group:https://groupprops.subwiki.org/wiki/Signed_symmetric_group. Could someone tell me what are the normal subgroups of signed symmetric groups? (I think they should be very similar like $S_n$) Thank you.

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    $\begingroup$ See A. Kerber, Representations of Permutation Groups I. Springer-Verlag (LNM 240), (1971). The signed symmetric group is a special case of the generalized symmetric group, which is discussed in Kerber's book. $\endgroup$ – Dietrich Burde Nov 19 '19 at 9:12
  • $\begingroup$ @ Dietrich Burde Thank you for your help. $\endgroup$ – Xiaosong Peng Nov 19 '19 at 9:42
  • $\begingroup$ The number of normal subgroups of $S_n$ for $n=0,1,2,3,4,\ge 5$ is $1,1,2,3,4,3$. For these values the number of proper normal subgroups is $0,0,1,2,3,2$; the number of nontrivial proper normal subgroups is $0,0,0,1,2,1$. $\endgroup$ – YCor Nov 19 '19 at 22:00
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In brief, in ATLAS notation, for $n \ge 5$ there are $9$ normal subgroups of the signed permutation group $2^n\!:\!S_n$.

These have structures $1$, $2$, $2^{n-1}$, $2^n$, $2^{n-1}\!\!:\!A_{n}$, $2^n\!:\!A_n$, $2^{n-1}\!\!:\!S_n$ (two subgroups) and $2^n\!:\!S_n$.

Note that the normal subgroup $2^{n-1}$ consists of diagonal matrices of determinant $1$ in the standard linear representation. Also $A_n$ acts irreducibly on $2^{n-1}$ when $n$ is odd, but when $n$ is even, the subgroup $2$ is contained in the subgroup $2^{n-1}$, with irreducible action of $A_n$ on the quotient of order $2^{n-2}$.

There are 11 normal subgroups when $n=4$, 9 when $n=3$, and 6 when $n=2$.

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  • $\begingroup$ Thank you. Could you tell me or provide some references about the concrete forms of these normal subgroups? $\endgroup$ – Xiaosong Peng Nov 19 '19 at 9:44
  • $\begingroup$ The notation describes the structure moderately accurately. I have added a little more explanation. $\endgroup$ – Derek Holt Nov 19 '19 at 10:06

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