Subgroups of $D_3$

Question

What are all the subgroups of $D_3$, I have so far if $D_3=\{r_0, r_1,r_2, s_1, s_2, s_3\} $ then the subgroups are $$\{r_0, r_1, r_2\},\{r_0, s_1\}, \{r_0, s_2\}, \{r_0, s_3\}, D_3, \{r_0\}. $$

Are there any more that I've missed?

Answer 1

The proper subgroups of a group with $6$ elements has $1,2$ or $3$ elements. Therefore all proper subgroups need to be cyclic and we need to check the order of elements.

Answer 2

You have found them all. To verify this, first note that we can view $D_3$ as a group which permutes the three vertices of a triangle, so it is (isomorphic to) a subgroup $H$ of $S_3$. Moreover, $D_3$ and $S_3$ both have six elements, so in fact $H$ must be all of $S_3$. Therefore analyzing $D_3$ is the same as analyzing $S_3$.

Now, since $S_3$ is not cyclic (it is not even abelian), it has no element of order $6$, so by Lagrange's theorem, all of its elements have order $1$, $2$, or $3$.

By writing the elements in cycle notation, we see that there are two elements of order $3$, namely $(123)$ and $(132)$, and these are inverses of each other, so they lie in a subgroup of order $3$, and this is the only subgroup of order $3$.

There are three elements of order $2$, namely $(12)$, $(13)$, and $(23)$, and each of these is in a distinct subgroup of order $2$, so there are three subgroups of order $2$.

Finally, of course, there are the trivial subgroup and the full group. We have covered all possible subgroup orders, so we're done.