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.)

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  • 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!

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