Center of a group

  • #1
389
0
"If N is a cyclic subgroup that is normal in G with index n and Aut(N) has no element of order n then N is central."

Is this true?

I think it is, but I don't know how to go about proving it... anyone have a hint that could get me started?
 

Answers and Replies

  • #2
mathwonk
Science Advisor
Homework Helper
2020 Award
11,267
1,470
well lets just follow our nose. if N is not in the center, then its elements do notcommute with everything. so conjugating N by other elements of G, altough it elaves N \inv ariant, does not always give the identity automorphism of N.

hence N has some automorhisms given by these other elements.

now what would the orders of such automorphisms be? surely since N is cyclic, elements of n itelf do act trivially on N by conjugation, so the conjugation action defines map from G to Aut(N) with N in the kernel.

Thus the image of G/N in Aut(N) has image of roder dividing indexN = n.

But it is not clear to me it must be cyclic, nor even of order n.

but suppose n were say prime? then what?

look for a counter example as a semidirect product of two cyclic groups.
 

Related Threads on Center of a group

  • Last Post
Replies
3
Views
2K
Replies
9
Views
273
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
14K
Replies
1
Views
837
  • Last Post
Replies
2
Views
725
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
Top