Cyclic groups and isomorphisms

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!
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?


Science Advisor
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. 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.

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