# Proof on Cyclic Subgroup Generation

So I am asked to assume that elements a and b of a group G have orders 14 and 22, respectively and that the intersection of $\langle a\rangle$ and $\langle b\rangle$ of the cyclic subgroups generated is not the trivial subgroup {1}. Prove that $a^7$ and $b^{11}$ is even.

I really don't know how to start this problem and am confused about cyclic groups. Can anyone help? And when you write it can you be detailed and explicitly highlight the concepts I need to understand? This is coming from someone who is taking group theory for the very first time.

Assumes that $<a>\cap<b>=<c>$.Order of $c$ can only be 2 because $\gcd(14,22) = 2$. Obviously $c$ is even and $c = a^7 = b^{11}$.
Since every cyclic group is abelian, I find it strange that you choose to write the identity element as $1$. I would prefer $0$ or $e$.
Working with the task at hand; you are given that $\langle a\rangle$ and $\langle b \rangle$ have nontrivial intersection. You are also given that they are of orders $14$ and $22$, respectively. We note that their only common divisor is $2$. Recall that Lagrange's theorem states that the order of a subgroup must divide the order of the group. As such, you know that their intersection is $\{ e, k \}$ for some $k$. Since $k$ is an element of both of these groups and $k^2 = e$, we have that $k = a^7 = b^{11}$. Do you see why?
• I don't quite understand the last part where $k=a^7=b^11$. But aside from that everything is fine. – cambelot Oct 4 '14 at 18:02
• Also how did you get $k^2$=e? – cambelot Oct 4 '14 at 18:12
• The intersection of two (sub)groups is a subgroup (prove this!). Since $\{k,e\}$ is a group, you must have $k^2 = e$. Recall that your group is abelian; this should help with intuition for your first question. – Andrew Thompson Oct 4 '14 at 20:43