# How to find the product of two subgroups of a group of units

In a group $$U(24)$$ = $$\{1,5,7,11,13,17,19,23\}$$, if $$H=\{1,13\}$$ and $$K=\{1,17\}$$ then HK is given by

1. $$\{1,3,1,17\}$$
2. $$\{1,13,17\}$$
3. $$\{1,5,13,17\}$$
4. $$\{1,13,17,221\}$$

I am preparing for an entrance exam and while solving previous year's question papers, I came to this question. I tried to find out how to solve this but I really could not. I am having no idea; how to solve this question. I know how to multiply two sets but I am confused about this one(subgroups). I can tell that this product is related to unitary groups somehow but I am having no idea at all to solve this. Please provide the solution, any help would be appreciated!

• I think you mean unit groups rather than unitary groups. Unitary groups are groups of matrices that preserve a Hermitian form. – Derek Holt 2 days ago

To multiply two subgroups, you take each element in the first subgroup and multiply it by each element in the second subgroup, and assemble all the answers in the result.

So you have: $$\{1 \times 1, 1 \times 17, 13 \times 1, 13 \times 17\}$$

Then you work out what each of the terms is:

$$\{1, 17, 13, 221\}$$

... but then you have to remember to take the modulus of each element mod $$24$$, so you need $$221 \pmod {24} = 5$$.

You know that $$13^2=17^2=1$$. Since the group is abelian, you only need to calculate $$13 \cdot 17 = 5$$. Therefore, the product $$HK$$ is the third option.
More explanation: $$HK$$ consists of elements $$13^n17^m$$ for integers $$n,m$$. But since $$13^{-1}=13$$ and $$17^{-1}=17$$, you only need to consider positive integers. And again, since $$13^2=17^2=1$$, you only need to consider the integers $$1$$ and $$2$$.