Thread: Finite Groups, Centre and Class Equation

Dummit and Foote Section 4.3 Groups Acting on Themselves by Conjugation - The Class Equation - Exercise 6 reads as follows:

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Assume G is a non-abelian group of order 15. Prove that Z(G) = 1. Use the fact that <g> (g) for all g G to show that there is at most one possible class equation for G

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Can anyone help me get started on this problem?

ok, since the class equation is:



and since Z(G) ≠ G, and ,

we have either:

15 = 1 + 3k + 5m
15 = 3 + 3k + 5m
15 = 5 + 3k + 5m

we want to show that only the first one is possible.

suppose 15 = 3 + 3k + 5m.

clearly m = 0,1 or 2.

now, this means that 15 - 5m is divisible by 3, so m = 0. this means that for all non-central g. so contains no elements of order 3, which is impossible since all of Z(G) is in .

on the other hand, suppose that:

15 = 5 + 3k + 5m.

then k = 0,1,2 or 3, so 15 - 3k is divisible by 5, so k = 0. then for every non-central g, which again leads to a contradiction.

Originally Posted by Bernhard

Dummit and Foote Section 4.3 Groups Acting on Themselves by Conjugation - The Class Equation - Exercise 6 reads as follows:

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Assume G is a non-abelian group of order 15. Prove that Z(G) = 1. Use the fact that <g> (g) for all g G to show that there is at most one possible class equation for G

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Can anyone help me get started on this problem?

The basic idea is this. From the class equation you know that if denotes the number of conjugacy classes of size and the number of size then . Now, if is non-abelian you know that and so . So what are the possibilities?

Thanks.

You write "since Z(G) G and 1< |G: (g)| <15

we have either:

15 = 1 + 3k + 5m
15 = 3 + 3k + 5m
15 = 5 + 3k + 5m

Can you be more explicit about why this follows - why the 3k + 5m?

Peter

Sorry - posted my next question before I read this ... just reflecting over your post now

Peter

Clarification on Drexel28's post:

Drexel28 writes |Z(G)| = 3, 5, 15 ... ... but Dummit and Foote imply in the problem that |Z(G)| = 1 since Z(G) = 1

But then how do we satisfy the equation 15 = |Z(G)| + 5x + 3y?

Can anyone help?

Peter

i believe that is a typo, he meant |Z(G)| = 1,3 or 5 (by assumption, G is non-abelian, so we can't have Z(G) = G, which is the case for every abelian group).

OK, yes

But if then given Z(G) = 1 we have 15 = |Z(G)| + 5x + 3y = 1 + 5x + 3y which does not solve for any x and y

Peter

oh? what about x = 1, and y = 3?

let's break that down even further. for an element g of order 3, this means that since its centralizer is of index 5, and we thus have just 2 elements of order 3.

it also means that for any element h of order 5, and we have 3 such centralizers, leading to 12 (3*4) elements of order 5. this gives us:

12 + 2 + 1 = 15 elements in all.

as a cruel aside, i remark that in point of fact, there actually aren't any non-abelian groups of order 15, and that any abelian group of order 15 is, in fact, cyclic. you already have the means to prove this fact, but an elementary proof is (as is often the case), not so easy to puzzle out. but you should try.

assume G is non-abelian. then Z(G) = {e}, and as we see, G has only 2 elements of order 3.

a) show the subgroup of order 3 is normal (hint: what order must any conjugate have?)
b) if y is an element of order 5, and x an element of order 3, write:

2 different ways, one using the order of y, and one considering x conjugated by y 5 times:



(from part (a) you have only two choices for a conjugate of x, what are they?)

derive a contradiction, and conclude therefore, that no such group exists.

Thanks so much Deveno

It is fascinating how much (if you are skilled enough) the theory enables you to conclude about a group of order 15!

Peter

Originally Posted by Deveno

oh? what about x = 1, and y = 3?

let's break that down even further. for an element g of order 3, this means that since its centralizer is of index 5, and we thus have just 2 elements of order 3.

it also means that for any element h of order 5, and we have 3 such centralizers, leading to 12 (3*4) elements of order 5. this gives us:

12 + 2 + 1 = 15 elements in all.

as a cruel aside, i remark that in point of fact, there actually aren't any non-abelian groups of order 15, and that any abelian group of order 15 is, in fact, cyclic. you already have the means to prove this fact, but an elementary proof is (as is often the case), not so easy to puzzle out. but you should try.

assume G is non-abelian. then Z(G) = {e}, and as we see, G has only 2 elements of order 3.

a) show the subgroup of order 3 is normal (hint: what order must any conjugate have?)
b) if y is an element of order 5, and x an element of order 3, write:

2 different ways, one using the order of y, and one considering x conjugated by y 5 times:



(from part (a) you have only two choices for a conjugate of x, what are they?)

derive a contradiction, and conclude therefore, that no such group exists.

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You write above:

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it also means that for any element h of order 5, and we have 3 such centralizers, leading to 12 (3*4) elements of order 5. this gives us:

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Why (how?) does it lead to 3*4 elements of order 5 ... where does the 4 come from?

Peter

in any group of order 5, there are 4 elements of order 5, and 1 element (e) of order 1. normally, when you count group elements in subgroups, you only count non-identity elements (since e is an element of EVERY subgroup, so we don't want to count it more than once).

so in a group of prime order, p, we have the element e, and p-1 elements of order p. if we have k such subgroups, the total number of elements in all of them together has to be 1 + k(p-1).

in this case, our prime is 5, and 5-1 = 4. that's where the 4 comes in. we're not counting e, because that's already accounted for as Z(G).

Originally Posted by Deveno

in any group of order 5, there are 4 elements of order 5, and 1 element (e) of order 1. normally, when you count group elements in subgroups, you only count non-identity elements (since e is an element of EVERY subgroup, so we don't want to count it more than once).

so in a group of prime order, p, we have the element e, and p-1 elements of order p. if we have k such subgroups, the total number of elements in all of them together has to be 1 + k(p-1).

in this case, our prime is 5, and 5-1 = 4. that's where the 4 comes in. we're not counting e, because that's already accounted for as Z(G).

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In the above you write:

"in any group of order 5, there are 4 elements of order 5, and 1 element (e) of order 1."

I am really surprised and intrigued by this. Can you explain how this comes about and/or give me a reference for the theory behind such a conclusion?

Peter

consider that for any group of order 5, the only possible orders for an element are 5 and 1, because 5 is a prime number. if an element is not the identity, it does not have order 1, therefore: it has order 5.

furthermore, since we have (several) elements of order 5, they are all generators for the group, hence it is cyclic.

for example: consider (Z5,+), the group of integers mod 5 under addition modulo 5:

clearly 1 has order 5.
2+2 = 4, 2+2+2 = 1, 2+2+2+2 = 3, 2+2+2+2+2 = 0, 2 has order 5 (in an additive group a^k is usually witten ka. this does not necessarily mean multiplication by k).
3+3 = 1, 3+3+3 = 4, 3+3+3+3 = 2, 3+3+3+3+3 = 0, 3 has order 5.
4+4 = 3, 4+4+4 = 2, 4+4+4+4 = 1, 4+4+4+4+4 = 0, 4 has order 5.

so 0 has order 1, and 1,2,3 and 4 all have order 5.

with a generic cyclic group of order 5, it is just the same:

e has order 1, and a, a^2, a^3, and a^4 all have order 5 (note that this just moves the addition modulo 5 of Z5 to the exponent of a).