Can $C_{10}$ be isomorphic to $C_5\times C_2$? $C_{10}$ is algebraically described by $a^{10}=1$. That's all. $C_{10}$ is $\{1,a,a^2,a^3,a^4,a^5,a^6,a^7,a^8,a^9\}.$
$C_5\times C_2$ is algebraically described by $r^5=1$, $f^2=1$, and $rf=fr$. $C_5\times C_2$ is $\{1,r,r^2,r^3,r^4,f,fr,fr^2,fr^3,fr^4\}$.
Prove or disprove $C_{10}$ isomorphic to $C_5\times C_2$.
My question is : when construct the operation $\phi$ I got they are isomorphic. However, to definition of isomorphic graphs, $C_{10}$ with 10 edges while $C_5\times C_2$ has 15 edges, they cannot be isomorphic. That fact makes me confused. Please, explain me what's wrong, what's right?
 A: Define a map $\phi : C_5 \times C_2 \to C_{10}$ by $\phi(r^u,f^v) = a^{2u + 5v}$.  Show now that $\phi$ is an isomorphism of groups. 
A: The two groups are indeed isomorphic. Since $C_{10}$ and $C_5\times C_2$ both have $10$ elements, it suffices to show that there is an element of order $10$ in $C_5\times C_2$ (i.e. a generator), because all cyclic groups of the same order are isomorphic. I claim that $rf$ is a generator of $C_5\times C_2$. This is not hard to verify, so once you show this, you have shown that $C_{10}\cong C_5\times C_2$, and that the desired isomorphism is
\begin{align*}
\phi : C_{10}&\to C_{5}\times C_2\\
a&\mapsto fr
\end{align*}
Also, I'll note that the presentation of a group is not necessarily unique. We can describe a group with more generators than necessary, and that's what you've done in the case of $C_5\times C_2$. Since it's isomorphic to $C_{10}$, it can be presented the same way as $C_{10}$, but the presentation with two generators, $r$ and $f$, is also legitimate. If you look in the graph you've constructed of $C_5\times C_2$, you should be able to look at the edges defined by multiplication by $rf$ and see that there are $10$ of them. You could also make the graph of $C_{10}$ differently by looking at two sets of edges defined by multiplication by $a^2$ and $a^5$, respectively. That graph should look like your graph of $C_5\times C_2$.
