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$G$ 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...

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1 Answers1

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Your proof is correct -- $\langle x, y \rangle$ must be the entire group.

I couldnt think of any example though...

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.

In general, the groups of order $2p$ for a prime $p$ are $C_{2p}$ and $\text{Dih}_p$.

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Caleb Stanford
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