Cyclic groups and isomorphisms

In summary, the conversation discusses the proof that G is isomorphic to C_3 x C_5 when G has order 15. The difficulty lies in proving this, but it can be shown using an "elementary" proof that doesn't use Sylow's theorem. This proof involves using Cauchy's theorem to find elements of order 3 and 5, defining an equivalence relationship on G, and showing that the size of the conjugacy class must divide the order of G. The conversation then explores the different possibilities for the sizes of the conjugacy classes and ultimately concludes that every group of order 15 is isomorphic to C_15, making G isomorphic to C_3 x C_5.
  • #1
blahblah8724
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I have a question where it says prove that [itex] G \cong C_3 \times C_5 [/itex] when G has order 15.

And I assumed that as 3 and 5 are co-prime then [itex] C_{15} \cong C_3 \times C_5 [/itex], which would mean that [itex] G \cong C_{15} [/itex]?

So every group of order 15 is isomorohic to a cyclic group of order 15?

Doesn't seem right?

Help would be appreciated! Thanks!
 
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  • #2
Yes, it is right. The only group of order 15 is [itex]C_{15}[/itex]. The difficulty is in actually proving this. Do you know Sylow's theorem?
 
  • #3
here is an "elementary" proof (one that doesn't use the Sylow theorems).

by cauchy's theorem, we have an element a of order 3, and an element b of order 5. if a and b commute, then ab is an element of order 15, so G is cyclic.

for x,x' in G, define x~x' if and only if there is some g with x' = gxg-1. this defines an equivalence relationship on G, and [x] is called the conjugacy class of x. note that any two conjugates must have the same order. if we define the subgroup

N(x) = {g in G: gx = xg}, we have the following bijection:

G/N(x) <--> [x] beween left cosets of N(x) and elements of [x],

given by hN(x) = hxh-1. to see this suppose hN(x) = h'N(x). then h'-1h is in N(x), so:

h'-1hx = xh'-1h
hx = h'xh'-1h
hxh-1 = h'xh'-1

so h and h' give rise to the same conjugate of x, and reversing the argument shows that distinct conjugates of x give rise to distinct (left) cosets of N(x).

this means that the SIZE of [x] is [G:N(x)], the index of N(x) in G, which in particular, MUST divide G.

now the size of [e] is 1, and since |G| = Σ|[x]| (since the conjugacy classes partition G), we have:

15 = k + 3m + 5n,

where k is the number of conjugacy classes of size 1, m is the number of conjugacy classes of size 3, and n is the number of conjugacy classes of size 5.

now k is at least 1, and if k > 1, then we have some element besides e that commutes with everything. if it's a, an element of order 3, then a commutes with b, an element of order 5, and thus all of G is abelian, and thus (as we saw above) G is cyclic. and if it's b, an element of order 5, then G is likewise abelian (and thus cyclic).

so the only case that might possibly be left is k = 1 (so that no element of order 3, and no element of order 5 commutes with everything). in which case:

14 = 3m + 5n.

11 is not divisible by 5, so m is not 1.
8 is not divisible by 5, so m is not 2.
5 is divisible by 5, so m might be 3.
2 is not divisible by 5, so m is not 4.

so the "bad case" is where k = 1, m = 3, n = 1.

so we must have one conjugacy class with 5 elements, and 3 conjugacy classes with 3 elements.

so let's look at the conjugacy class of b, . two elements of are aba-1, and a-1ba. suppose aba-1 was a power of b:

aba-1 = bk

then b = a3ba-3 (since a3 = e)
= a2(aba-1)a-2
= a2bka-2
= a(abka-1)a-1
= a(aba-1)ka-1 = a(bk)ka-1
= a(bk*k)a-1 = (aba-1)k*k
= bk*k*k

that is k3 = 1 (mod 5), which means k = 1, and thus a and b commute, and G is abelian, and thus cyclic.

otherwise, <aba-1> is a different subgroup of order 5 than <b>. in the same way, if G is non-abelian, then <a-1ba> is a different subgroup than <b>.

now if <aba-1> = <a-1ba>, then:

a-1ba = (aba-1)j = abja-1
ba = a2bja-1
b = a2bja-2
ab = bja-2
aba-1 = bj,

contradicting our assumption that <aba-1> ≠ <b>.

but...this gives us 12 distinct elements of order 5, which means we have just 2 elements of order 3. since by supposition, these two elements must be conjugate (we are assuming no non-identity element commutes with everything), we must have a conjugacy class with just two elements, which is impossible, since 2 does not divide 15.

so...in all POSSIBLE cases, there exists some element of order 3 that commutes with an element of order 5, and thus G is abelian, and therefore cyclic.
 

1. What is a cyclic group?

A cyclic group is a mathematical group that is generated by a single element, called a generator, through repeated application of its group operation. This means that all the elements of the group can be obtained by taking the generator and combining it with itself a certain number of times.

2. How are cyclic groups and isomorphisms related?

Cyclic groups and isomorphisms are related because isomorphisms preserve the structure and properties of a group. This means that if two groups are isomorphic, then their cyclic subgroups will also be isomorphic. In other words, if two groups have the same structure, their cyclic subgroups will also have the same structure.

3. What is the order of a cyclic group?

The order of a cyclic group is the number of elements in the group. Since a cyclic group is generated by a single element, its order is equal to the order of the generator. For example, a cyclic group generated by an element of order 3 will have 3 elements.

4. How can I determine if two cyclic groups are isomorphic?

To determine if two cyclic groups are isomorphic, you can compare their orders and see if they are equal. If the orders are equal, then the two groups are isomorphic. You can also look for a generator in each group and see if they produce the same elements when combined with themselves a certain number of times.

5. Can a non-cyclic group be isomorphic to a cyclic group?

Yes, a non-cyclic group can be isomorphic to a cyclic group. Isomorphisms only require that the two groups have the same structure, so it is possible for a non-cyclic group to have a subgroup that is isomorphic to a cyclic group.

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