Proving Normal Subgroup N is Central in Group G

In summary, the conversation discusses the statement "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." The participants consider whether this statement is true and attempt to prove it. They also discuss the possibility of finding a counter example, using the example of a semidirect product of two cyclic groups.
  • #1
Cincinnatus
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?
 
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  • #2
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.
 
  • #3


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
 

Related to Proving Normal Subgroup N is Central in Group G

1. What does it mean for a subgroup to be central in a group?

For a subgroup N to be central in a group G, it means that all elements of N commute with every element of G. In other words, for every element n in N and every element g in G, the product ng=gn. This is denoted as [N,G] = {ng | n∈N, g∈G}={gn | n∈N, g∈G}.

2. How can I prove that a subgroup N is central in a group G?

One way to prove this is by showing that every element in N commutes with every element in G. This can be done by directly calculating the product of every pair of elements in N and G, or by using the definition of a central subgroup and proving that [N,G] = {e}, where e is the identity element of G.

3. Can a subgroup be both central and normal in a group?

Yes, a subgroup can be both central and normal in a group. A subgroup that is both central and normal is called a central subgroup.

4. Is there a specific method or algorithm for proving that a subgroup is central in a group?

There is no specific method or algorithm for proving that a subgroup is central in a group. It is usually done by directly calculating the product of every pair of elements in the subgroup and the group, or by using the definition of a central subgroup and proving that [N,G] = {e}.

5. What are some applications of proving that a subgroup is central in a group?

Proving that a subgroup is central in a group can be useful in various fields of mathematics, such as group theory, algebra, and geometry. It can also be used in cryptography, as central subgroups play an important role in constructing secure encryption algorithms. Additionally, understanding central subgroups can help in the study of group structures and their properties.

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