A comment on a proof of equivalent knots The following theorem and its proof is from A First Course in Algebraic Topology
 By Czes Kosniowski pp. 219-220 http://books.google.jo/books?id=vvU3AAAAIAAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false
I have only one comment on the following proof. In the last two lines: $\mathbb Z_m \times \mathbb Z_n \cong \mathbb Z_{m'}\times \mathbb Z_{n'}$ iff $\{n,m\}=\{n',m'\}$ because $gcd (n.m)=1$, and $gcd(m',n')=1$. I don't think that is precise!! 
Do you know how last statement in the proof should be in order to have correct proof?





 A: Apply the classification of finite abelian groups http://torus.math.uiuc.edu/jms/m317/handouts/finabel.pdf
A: Well, you're right, it's not true as stated. (I mean the argument in the last sentence; the result itself is, of course, true.)  For example, take $m = 6, n = 5$ and $m^{\prime} = 3, n^{\prime} = 10$.  Then $\mathbb{Z}_{m}\times\mathbb{Z}_{n}\simeq \mathbb{Z}_{30}\simeq\mathbb{Z}_{n^{\prime}}\times\mathbb{Z}_{m^{\prime}}$. But you can get from $\langle x\mid x^m\rangle\ast\langle y\mid y^n\rangle\simeq\langle u\mid u^{m^{\prime}}\rangle\ast\langle v\mid v^{n^{\prime}}\rangle$ to $\{m,n\} = \{m^{\prime},n^{\prime}\}$ by properties of the free product.
Let's identify the isomorphic groups $\langle x,y\mid x^m, y^n\rangle = \langle u,v\mid u^{m^{\prime}}, v^{n^{\prime}}\rangle$. Now, $u$ is an element of finite order $m^{\prime}$, so it is conjugate to an element of $\langle x\rangle$ or of $\langle y\rangle$.  In particular, the order $m^{\prime}$ of $u$ divides either $m$ or $n$ and, since $m$ and $n$ are relatively prime, it divides exactly one of them. Say, without loss of generality, that $m^{\prime}$ divides $m$.  Similarly, $v$ is conjugate to a power of either $x$ or $y$ so, by coprimality again, $v$ is a power of $y$, and so its order $n^{\prime}$ divides $n$. By symmetry, we also have that $m$ divides $m^{\prime}$ and $n$ divides $n^{\prime}$. Thus, $m = m^{\prime}$ and $n = n^{\prime}$.
