# Is it true that if $g,h\in G$ have order $p$, where $p$ is prime, the only possible order of $\langle g\rangle \cap \langle h \rangle$ is $p$?

Is it true that if $g,h\in G$ have order $p$, where $p$ is prime, the only possible order of $\langle g\rangle \cap \langle h \rangle$ is $p$?

Here is my attempted proof of the result. Is it correct?

Since $o(g) = o(h) = p$, which is finite, it follows that $\langle g\rangle, \langle h\rangle$ are cyclic groups of the same order, and since $\left( \langle g\rangle \cap \langle h \rangle \right) ≤ \langle g\rangle$, $\langle g\rangle \cap \langle h \rangle$ is also cyclic. So, disregarding the identity element (trivally with order $1$), $\forall x\in \left( \langle g\rangle \cap \langle h \rangle \right), \text{ }o(x)\mid\lvert\ \langle g\rangle \cap \langle h \rangle \rvert$. But $o(x) = o(g) = o(h) = p$ so $p\mid \lvert\ \langle g\rangle \cap \langle h \rangle \rvert$.

But $\lvert \langle g \rangle \rvert = p$ and since $\langle g\rangle \cap \langle h \rangle$ is a subgroup of $\langle g \rangle$, by Lagrange's Theorem $\lvert\ \langle g\rangle \cap \langle h \rangle \rvert \mid p$.

So $\lvert\ \langle g\rangle \cap \langle h \rangle \rvert = p$ and we are done.

I'm not sure about my $o(x) = p$ step, however; it's intuitively true, but how could I go about proving this?

• Well, it could be $1$. Apr 20, 2015 at 12:17
• Applying Lagrange is enough, no need to bother which of these is cyclic (of course they all are) Apr 20, 2015 at 12:20
• Okay so other than the silly mistake that was overlooking the case $\ \langle g\rangle \cap \langle h \rangle = e$, along with the added extra fluff in the proof, was this correct? Apr 20, 2015 at 12:25
• $$o(x) \mid \lvert \cdots \rvert$$ may not be as bad as the legendary $$\Xi \over \bar\Xi$$ but I do still think that it deserves a good old "your notation sucks!" May I suggest using $\#$ for "number of elements of" next time? Apr 20, 2015 at 12:26
• Or better yet: "The order of $x$ divides the order of $\langle g \rangle \cap \langle h \rangle$." Apr 20, 2015 at 13:58

Yes, it is correct that $o(x) = p$. Indeed, thanks to Langrange's theorem you can prove that every non-identity element in a group $G$ with $\#G = p$ has order $p$ (just consider the subgroup generated by that element).
As Hagen von Eitzen suggested in his comment, it is not necessary to say this explicitly, though. Indeed $$\langle g \rangle \cap \langle h \rangle \leq \langle g \rangle$$ therefore by Lagrange's theorem $$\#(\langle g \rangle \cap \langle h \rangle) \mid \# \langle g \rangle = p$$ so $\#(\langle g \rangle \cap \langle h \rangle)$ is either $1$ or $p$.