Find the number of elements of $\Bbb Z_{25}\times \Bbb Z_5$ of order $5.$

Question

Find the number of elements of $\Bbb Z_{25}\times\Bbb Z_5$ of order $5.$

My solution goes like this:

For $\Bbb Z_{25}\times \Bbb Z_5$, we count $(a,b)\in\Bbb Z_{25}\times\Bbb Z_5$, such that $o((a,b))={\rm lcm}(o(a),o(b))=5.$

• Case 1: If $o(a)=o(b)=5$, then $\Bbb Z_{25}$ and $\Bbb Z_{5}$ has an element of order $5$ each by Cauchy's theorem for abelian groups. Now, each such element of order $5$ generates a cyclic subgroup of order $5.$ Also, each element in the cyclic group apart from the identity element is of order $5$. Thus, $a$ and $b$ has $4$ possible choices each. Thus, we have $16$ total elements in this case.

• Case 2: If $o(a)=1,o(b)=5$, then we have a cyclic group of order $5$ generated by $b.$ Thus, we have $b$ has $4$ possible choices.

• Case 3: If $o(a)=5,o(b)=1$. Similarly, like Case 2, we have $4$ possible choices of $a$.

So, we have $24$ elements of order $5.$

Is the above solution, correct? If not, where is it going wrong?

No, this link Find the number of elements of order $p$, $p^2$, $p^3$, respectively in the group $\mathbb Z/p^3\mathbb Z \times \mathbb Z/p^2\mathbb Z$ does not answer my question as my question focuses upon a proof verification /solution verification but that link is for the following question :"Find the number of elements of order $p$, $p^2$, $p^3$, respectively in the group $\mathbb Z/p^3\mathbb Z \times \mathbb Z/p^2\mathbb Z$ do..." which is not related to my query of the correctness of my approach as a solution to a problem. Furthermore, the user asks for a solution while I posted this problem for verifying my solution, I already have.

Answer 1

Your solution is correct, but I might consider it a bit incomplete. Specifically, it's not clear from your explanation of Case $1$ how you conclude that there are exactly $4$ choices for $a$. Your reasoning is that $\Bbb Z_{25}$ has some element of order $5$, which generates a cyclic subgroup. But how do you know the non-identity elements of that cyclic subgroup are the only elements of order $5$ in $\Bbb Z_{25}$? The same issue arises for Case $3$.

Here's another solution: Define the map: $\varphi: G \to G$ via $\varphi(x)=x^5$. It's easy to see (since $G$ is abelian) that $\varphi$ is a homomorphism and that $\varphi (G) \cong \Bbb Z_{25} / \Bbb Z_5 \times \Bbb Z_5/ \Bbb Z_5 \cong \Bbb Z_5$, so $\vert \ker \varphi \vert = \frac{125}{5}= 25$. A single element of the kernel has order $1$, and the rest have order $5$, so $G$ has $24$ elements of order $5$.