# The order of subgroup generated by two distinct elements of order 2

$G$ is a group of order $2$6. If $x$ and $y$ are two distinct elements of order $2$, what could the order of $\langle x,y\rangle$ be?

By Lagrange's theorem, $\langle x\rangle$ and $\langle y\rangle$ are subgroups of $\langle x,y\rangle$, so order $\langle x,y\rangle$ has to be divisible by $2$. $\langle x,y\rangle$ is a subgroup of $G$, so the possibilities are $2$, $13$, $26$. Since $13$ is not divisible by $2$, and $x,y$ have order $2$, so $x,y$ are not identities and hence order of $\langle x,y\rangle = 26$.

Is something wrong with my proof? I couldnt think of any example though...

• your proof seems okay to me – happymath Mar 5 '15 at 3:34
• It is correct. ${}\qquad{}$ – Michael Hardy Mar 5 '15 at 3:45
• It is good. One example is the dihedral group of order 26; here $x$ and $y$ could be any two reflections. – Brent Kerby Mar 5 '15 at 4:16

Your proof is correct -- $\langle x, y \rangle$ must be the entire group.
There are only two groups of order $26$ -- the cylcic group $C_{26}$, and the dihedral group $\text{Dih}_{13}$. $C_{26}$ has only one element of order $2$, so it won't be an example. So the only example is $\text{Dih}_{13}$, the group of symmetries of a $13$-gon. In this group, $x$ and $y$ are two different reflections of the $13$-gon; what you have proven is that two different reflections are enough to generate the entire set of symmetries.