Is Every Group of Order 15 Cyclic?

[blahblah8724]

I have a question where it says prove that $G \cong C_3 \times C_5$ when G has order 15.

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

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

Doesn't seem right?

Help would be appreciated! Thanks!

 

[micromass]

Yes, it is right. The only group of order 15 is $C_{15}$. The difficulty is in actually proving this. Do you know Sylow's theorem?

 

[Deveno]

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 = b^k

then b = a³ba^-3 (since a³ = e)
= a²(aba^-1)a^-2
= a²b^ka^-2
= a(ab^ka^-1)a^-1
= a(aba^-1)^ka^-1 = a(b^k)^ka^-1
= a(b^k*k)a^-1 = (aba^-1)^k*k
= b^k*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 = ab^ja^-1
ba = a²b^ja^-1
b = a²b^ja^-2
ab = b^ja^-2
aba^-1 = b^j,

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.