Let and be positive integers and . Let and . Let be the central product of and with an element of order identified; this group has the presentation . Describe as a direct product of two cyclic groups.

[Proof suggested by Fred H]

Note that and are relatively prime. By Bezout’s identity, there exist such that . Let .

Now . (For the second equality,we used the fact that and commute.) Similarly, . In particular, generates . By Lagrange’s Theorem, ; so .

We can write as a direct product of cyclic groups as either or . (Since, for instance, and are relatively prime.)

## Comments

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?

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!

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