# Why is the order of the subgroup 3?

I want to find the order of the subgroup $\langle ab\rangle$ of $D_3=\langle a,b\mid a^3=1,b^2=1,ba=a^2b\rangle$

According to my notes, the order of this subgroup is 3. But why is it like that?

I thought that it would be 2,because :

$$(ab)^1=ab$$ $$(ab)^2=(ab)(ab)=a(ba)b=aa^2bb=a^3b^2=1$$

Isn't it like that?

EDIT: The diagram of the subgroups of $D_3$ that is in my notes is:

But, is the right one maybe like that?

• Yes. You got it right. Jun 14 '14 at 14:33
• Are you sure it didn't say index rather than order? Jun 14 '14 at 14:34
• The index of $H$ in $G$ is the number of cosets $gH$. Jun 14 '14 at 14:37
• In general, if $H$ is a subgroup of a finite group $G$, then the index of $H$ in $G$ is denoted by $[G : H]$ and is given by $[G:H] = \frac{|G|}{|H|}$. Jun 14 '14 at 14:40
• The red numbers are intended to be the order (because the trivial subgroup has a red $1$ and the whole group has a red $6$). There are two errors in the diagram: the one you found and the analogous one for $\langle a^2b\rangle$. Jun 14 '14 at 14:50

As has been discussed in the comments, the subgroup $\langle ab\rangle < D_3$ has order two (as you correctly demonstrated) and has index three. Furthermore, your second diagram is correct; it is the subgroup lattice for $D_3$ where the red numbers indicate the order of each subgroup in the corresponding row.