- #1

- 389

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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?

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- Thread starter Cincinnatus
- Start date

- #1

- 389

- 0

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?

- #2

mathwonk

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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.

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