# Number of non-isomorphic groups of order 6

In the first part of the question, I showed that every group of order less than 6 is Abelian.

In the second part of the question I am asked to show that there are exactly 2 non-isomorphic groups of order 6.

First, what does that mean? Do I have to find two specific groups that are non-isomorphic to each other? Does it make sense to say that a group is non-isomorphic without saying what is it non-isomorphic to? And what does it have to do with the first part of the question?

Many thanks!

• It means "there are exactly two isomorphism classes of groups"; i.e. it wants you to find two groups $G_1$ and $G_2$ such that given any group $G$ of order $6$, exactly one of $G_1\cong G$ or $G_2\cong G$ holds. What is one group of order $6$ that you know? You said that every group of order less than $6$ is Abelian, so what about of order exactly $6$? – Hayden Jun 11 '15 at 1:01

The point is that you want to find two groups of order $6$ which aren't isomorphic to each other and then show that any other group of order $6$ must be isomorphic to one of those. For any given order, we know that there is at least one group of that order (the cyclic group of that order). With that in mind, we know that one of them is $\Bbb Z_6$. You might be tempted to guess what the other group is but this isn't advisable. Besides cyclic groups, there is another class of finite groups that you should know. The hard part is showing that any other group is isomorphic to one of these.