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**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 $\displaystyle |C_G(g)| = 3$ since its centralizer is of index 5, and we thus have just 2 elements of order 3.

it also means that $\displaystyle |C_G(h)| = 5$ 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:

$\displaystyle y^5xy^{-5}$ 2 different ways, one using the order of y, and one considering x conjugated by y 5 times:

$\displaystyle x \to yxy^{-1} \to y(yxy^{-1})y^{-1} \to \dots$

(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.