Abelian group and its orders Problem: Suppose an abelian group has an element $a$ of order $4$ and an element $b$ of order $3$. Show that it must also have elements of order $2$ and $6$.
Attempt: Since the group is abelian, $ab = ba$. 
Let $G = \{a, b, c, d\}$, where $a^4 = e$, $b^3 = e$, and I will assume $c^2 = e$ and $d^6 = e$.
Since the group is abelian, the order of ab will be the product of its order, so $(ab)^{12} = e$. If we try to find the order of $(ac)$ then $(ac)^8 = (a^8)(c^8) = (a^4)^2(c^2)^4$. So, we can see at least one of the elements has order 2 since.
Am I in the right track? Thank you, any help will be appreciated.
 A: let $a \in G$ (abelian group) s.t. $a^4=1$ and $b^3=1$. It's clear that $a \neq b$ and $ab$ has order $12$. what can you say about $(ab)^6$? and about $(ab)^2$?
obviously abelian property of $G$ is fundamental in this proof.
A: You starting setting things up right by choosing $a$ and $b$ to have orders 4 and 3 respectively. However, from there on you starting assuming the conclusion, which isn't OK.
Let us first summarise everything that we know.
We know that $a^4=e$, $b^3=e$, and that $a,a^2,a^3, b, b^2 \neq e$.
There's one important subtlety that I need to point out. We don't know that $a \neq b$ yet! Or  that $a^2 \neq b$. I suggest that you start off by proving that neither $a$ nor $b$ is a power of the other. 
Once you've proved that, we can construct a total of 12 different elements using $a$ and $b$ (Can you see why we can only construct 12 elements? Here, we're using the fact that $G$ is abelian.)
Try writing out all of these 12 different elements, and figuring out the order of each of them. Doing so will probably be enough for you to solve your problem!
However, once you've done so, maybe you could take a second or two to look back over this approach - was there any way you could have found these elements without making an exhaustive list? 
