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Transitive normality is not centralizer-closed

Revision as of 21:40, 16 February 2009 by Vipul (talk | contribs) (New page: {{subgroup metaproperty dissatisfaction| property = transitively normal subgroup| metaproperty = centralizer-closed subgroup property}} ==Statement== ===Statement with symbols=== Suppos...)
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Statement

Statement with symbols

Suppose H is a transitively normal subgroup of a group G. Then, it is not necessary that C_G(H) (the centralizer of H in G) is transitively normal.

Related facts

Related centralizer-closed properties

Related metaproperty dissatisfactions for transitively normal subgroups

Proof

Further information: symmetric group:S3

Suppose A is a cyclic group of order three, B is a symmetric group of degree three, and C is the subgroup of order three in B. Define:

G = A \times B, \qquad H = 1 \times C.

Then, H is a normal subgroup of prime order, hence is transitively normal. On the other hand, the centralizer of H in G is K = A \times C, which is not transitively normal. To see this, let \sigma be an isomorphism from A to C. Consider:

L := \{ (a, \sigma(a)) \mid a \in A \}.

L is a normal subgroup of K (since K is abelian). On the other hand, conjugating an element of L by an element of 1 \times B \setminus C sends (a,\sigma(a)) to (a,\sigma(a)^2), which is not in L. Hence, L is not normal in G. Thus, K is not transitively normal in G.