Proving Normal Subgroup N is Central in Group G

[Cincinnatus]

"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?

 

[mathwonk]

well let's 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.

 

[tacman]

Yes, this statement is true. To prove it, we can use the definition of a central subgroup, which states that a subgroup N of a group G is central if for all elements g in G, we have that gn = ng.

First, let's consider the case where N is a cyclic subgroup of G. This means that N is generated by a single element, say a. Since N is normal in G, we know that for any element g in G, we have gag^-1 = a^k for some integer k. This is because normality implies that conjugation by any element of G leaves N invariant.

Now, let's assume that Aut(N) has no element of order n. This means that for any element f in Aut(N), we have that fn ≠ e, where e is the identity element of N. From this, we can deduce that a^k ≠ e for any integer k ≠ 0 (since any element of Aut(N) can be written as a^k for some k).

Next, let's consider the element a^n in N. Since a is a generator of N, we know that a^n = e, the identity element of N. However, since Aut(N) has no element of order n, we have that (a^n)^k ≠ e for any integer k ≠ 0. This means that a^n = e is only possible if k = 0, which implies that a^n = e for all k in Z.

Now, let's take an arbitrary element g in G and consider the element gag^-1. We can write gag^-1 = a^m for some integer m, since N is cyclic. By the definition of a central subgroup, we know that gag^-1 = ag^-1g = a^m. This implies that g^-1ag = a^m, which can be rewritten as (g^-1ag)^n = a^(mn). Since we know that a^n = e, we have that (g^-1ag)^n = e.

But we also know that (g^-1ag)^n = g^-1a^n g = g^-1e g = e, since N is normal in G. This means that a^(mn) = e, which implies that mn = 0 (since a^n = e only for n = 0). But we also know that a^k ≠ e for any