We know that if a group is Abelian, then all its subgroups are normal. Also, if a group is nonabelian, it can contain a subgroup which is Abelian. Eg: The Dihedral group of order 2n, $D_{2n}$ is nonabelian for $n\geq 3$ but has $C_n$ sitting inside it. Now, my question is: "If a finite group is nonabelian, can it have a normal subgroup which is nonabelian?" In other words, can a normal subgroup of a nonabelian finite group be nonabelian? Thanks.
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1$\begingroup$ Yes, it can. Consider semidirect products, see here for $A_5\rtimes A_5$. $\endgroup$ – Dietrich Burde Feb 17 '15 at 19:39
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$\begingroup$ @Chuks I wonder why you think it couldn't. I would think that at least it'd be mentioned that if a subgroup is normal then it is abelian... $\endgroup$ – Timbuc Feb 17 '15 at 19:41
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2$\begingroup$ $A_n$ is normal in $S_n$ (index $2$) and for $n\geq 4$, $A_n$ is not abelian. $\endgroup$ – Mark Greer Feb 17 '15 at 19:44
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$\begingroup$ $D_{4n}$ has $D_{2n}$ as subgroup of index $2$ and subgroups of order $2$ are normal. Even easier is to take the direct product of two non-abelian groups. $\endgroup$ – j.p. Feb 19 '15 at 13:42
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Yes, the most trivial example would be to consider the nonabelian group itself (which is a normal subgroup). If you are looking for a proper normal subgroup, then consider $A_4$ as a normal subgroup of $S_4$.
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$\begingroup$ Thanks Uncountable et al. Is there a way of constructing finite groups that can have a particular normal subgroup. For example, I want to know all groups that has $D_8$ as a normal subgroup. $\endgroup$ – Chuks Feb 17 '15 at 19:52
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1$\begingroup$ We can take the product group $D_8\times G$ for any group $G$. Then $D_8\times \{e\}$ is a normal subgroup which is isomorphic to $D_8$. Other examples are $D_8$, $D_{16}$, $D_{24}$ etc.. $\endgroup$ – Uncountable Feb 17 '15 at 19:58
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$\begingroup$ Thanks for your attempt...but from your example, $D_{16}$ does not have $D_8$ as a proper normal subgroup. $\endgroup$ – Chuks Feb 17 '15 at 20:05
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1$\begingroup$ $\{e,r^2,r^4,\cdots,r^{14},s,sr^2,\cdots,sr^{14}\}$ is a normal subgroup of $D_{16}$ isomorphic to $D_8$. Note that I use the word $\textbf{isomorphic}$ here. I am not claiming that $\{e,r,\cdots,r^7,s,sr,\cdots sr^7\}$ is a subgroup of $D_{16}$, let alone that it is a normal subgroup. $\endgroup$ – Uncountable Feb 17 '15 at 20:07
