# Find easy isomorphic groups

21. December 2020 ‐ **2 min read**

Take a look at the following question:

Find a group isomorphic to $\langle \mathbb{Z}_{12}; + \rangle$ other than $\mathbb{Z}_{12}$. (FS19, 3a)

Let us investigate how we can provide a fast and simple answer here. The important theorem to have in mind here is *“A cyclic group of order n is isomorphic to $\langle \mathbb{Z}_n; \oplus \rangle$”* (Theorem 5.7)

In our specific example, this means that we have to find a cyclic group of order $12$. As WolframAlpha will tell you, there are quite a few such groups, but the easiest type are direct products of smaller groups (so $\mathbb{Z}_{n} \times \mathbb{Z}_{m}$). Since $\mathbb{Z}_n \times \mathbb{Z}_m$ has order $\text{lcm}(n,m)$, we need to find $n, m$ such that $\text{lcm}(n,m) = 12$ which is the same as $n \cdot m = 12$ if $\gcd(n,m) = 1$. For $12$, this would be $3 \cdot 4 = 12$.

Hence a simple solution to this question would be $\mathbb{Z}_{3} \times \mathbb{Z}_{4}$. (If you want to be more formal, also include the associated operation, so $\langle \mathbb{Z}_{3}; \oplus \rangle \times \langle \mathbb{Z}_{4}; \oplus \rangle$)

**Practice:** If you understood this, it should now be easy for you find a group isomorphic to $\langle \mathbb{Z}_{20}; + \rangle$ other than $\mathbb{Z}_{20}$.