Give three examples of groups of order 120, no two of which are isomorphic.

Question

Give three examples of groups of order 120, no two of which are isomorphic. Explain why they are not isomorphic. (This is Exercise 57 from Chapter 6 of the book Contemporary Abstract Algebra by Gallian.)

My Attempt:

$\mathbb Z_{120} $

$\mathbb Z_{2}\times \mathbb Z_{60}$

$\mathbb Z_{2}\times \mathbb Z_{2}\times \mathbb Z_{30} $

These are group of order 120, but I don't know if these are the isomorphic or not. Can any one help me?

Answer 1

That is correct!

One can show that the order of an element $(g_1, g_2,..., g_n)$ in a direct product of groups $G_1 \times G_2 \times \cdots \times G_n$ is $\operatorname{lcm}(|g_1|, |g_2|, ..., |g_n|)$.

Given this, think about $\max \{ \operatorname{ord}(g) \ | \ g \in G\}$ for each of the groups you've listed (orders would be preserved under isomorphism).

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Alternatively, there are small nonabelian groups of orders dividing $120$, allowing you to construct nonabelian groups of order $120$ via a direct product (abelian-ness should be preserved under isomorphism). You can even get a nonabelian group of order $120$ directly by considering the symmetry group of an $n$-gon for appropriately-sized $n$.

Answer 2

If there were an isomorphism, all orders of elements would be preserved. Does the second group have an element of order 120? How many elements of order 2 are there in each?