It is well known that characteristic subgroups of a group $G$ are normal. Is the converse true?
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The converse is wrong. Two counterexamples :
1) Let $G$ be the finite group of quaternions (of order $8$). Each subgroup of $G$ of index $2$ is normal but not characteristic.
2) Let $G$ be a vector space. Since $G$ is commutative, each of his subspaces are normal subgroups, but any non-trivial one is clearly not characteristic.
Louis La Brocante
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I think the simplest example is the Klein four-group, which you can think of as a direct sum of two cyclic groups of order two, say $A\oplus B$. Because it is abelian, all of its subgroups are normal. However, there is an automorphism which interchanges the two direct summands $A$ and $B$, which shows that $A$ (and $B$) are normal, but not characteristic. (In fact, the other non-trivial proper subgroup, generated by the product of the generators of $A$ and $B$ also works.)
James
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No. Wiki gives $S_3 \times C_2$ ($C_2$ being the cycling group of order 2) as an example--the center isn't a characteristic subgroup.
hoyland
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7The centre of any group is characteristic! The first factor $S_3$ in the direct product is normal but not characteristic. – Derek Holt Aug 17 '11 at 22:22
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You missed the word “fully” in that certainly confusing Wikipedia article. – Carsten S Sep 17 '21 at 08:06