## Describe a given central product as a direct product

Let $m$ and $n$ be positive integers and $d = \mathsf{gcd}(m,n)$. Let $Z_m = \langle x \rangle$ and $Z_n = \langle y \rangle$. Let $A$ be the central product of $Z_m$ and $Z_n$ with an element of order $d$ identified; this group has the presentation $A = \langle x,y \ |\ x^m = y^n = 1, xy = yx, x^{m/d} = y^{n/d} \rangle$. Describe $A$ as a direct product of two cyclic groups.

[Proof suggested by Fred H]

Note that $m/d$ and $n/d$ are relatively prime. By Bezout’s identity, there exist $a,b \in \mathbb{Z}$ such that $a(m/d) + b(n/d) = 1$. Let $z = x^by^a$.

Now $z^{n/d} = (x^by^a)^{n/d}$ $= (x^b)^{n/d}(y^a)^{n/d}$ $= (x^{b(n/d)} (y^{n/d})^a$ $= x^{b(n/d)} (x^{m/d})^a$ $= x^{b(n/d) + a(m/d)}$ $x^1 = x$. (For the second equality,we used the fact that $x$ and $y$ commute.) Similarly, $z^{m/d} = y$. In particular, $z$ generates $A$. By Lagrange’s Theorem, $|A| = mn/d = \mathsf{lcm}(m,n)$; so $A \cong Z_{\mathsf{lcm}(m,n)}$.

We can write $A$ as a direct product of cyclic groups as either $Z_{m/d} \times Z_n$ or $Z_m \times Z_{n/d}$. (Since, for instance, $m$ and $n/d$ are relatively prime.)

• Fred H  On June 8, 2011 at 11:13 am

I like this blog; you’ve helped me work through several tricky problems from D&F. Here, however, you’ve let me down :-).

Typos aside, psi is clearly not an automorphism: alpha^(m/d) is the identity, but x^(m/d) is z. Also, at the end you seem to assume that d is coprime to m/d and n/d, which is not true in general.

It seems to me that the resulting group is cyclic in all cases, so I was puzzled by the instruction to write it as a product. Here’s the idea: Let x,y be integers such that x(m/d) + y(n/d) = 1, and let g = a^y b^x. Then g^(n/d) = a and g^(m/d) = b, so the order of g is a common multiple of m and n. Since |A| = lcm(m,n), g must be a generator. Have I messed up too?

• nbloomf  On June 8, 2011 at 7:49 pm

Well I hate to disappoint. I’ll have my people send a refund. 🙂

Your solution looks reasonable. I’ll think about it for a bit to make sure I understand it.

Thanks for pointing out the mistake and suggesting a fix!

• nbloomf  On June 8, 2011 at 8:15 pm

Alright- I think I get it. That’s a slick proof. Thanks again!