Suppose an abelian group has an element m of order 4 and an element n of order 3. Show that it must also have elements of order 2 and 6. I am trying to manipulate $m^4$ and $n^3$ by setting them both equal to the identity e. I thought that squaring both sides of n^3=e would result in n^6=e. Thus the element n has order 6. I made a similar argument with m^4=e, but instead square rooted both side to get m^2=e. Thus showing the element m has order of 2. As both elements are in G and both orders required 2 and 6 are shown, I thought that would show it. Outside of my current thought process I don't know where to go with the problem though. If someone could lead me in the right direction I would appreciate it.
 A: If $m$ has order $4$, then no smaller power of $m$ is equal to the identity.  Therefore $m^2 \neq e$ but $(m^2)^2=m^4=1$.
The order of $mn$ is $12$ because $(mn)^k=m^kn^k$, and $m^kn^k \neq e$ for $k \lt 12$.  Using the same technique, then $((mn)^2)^6=e$.
More generally, in an abelian group, once you have an element of order $k$, you can find an element having order $d$ for any divisor $d\vert k$.  In fact, as pointed out in the comment below, you don't even need the group to be abelian for this to hold.
A: You have misunderstood the defintion of order of an element in a group. The order of an element $g$ in group $G$ with identity $e$, is the smallest integer $n$, such that $g^n=e$.
Now suppose an element $g\in G$ has even order, $2k$, i.e. $2k$ is the smallest integer $n$ such that $g^{n}=e$. Then $(g^2)^k=e$. Also if there is $l<k$, such that $(g^2)^l=e$, then $2l$ is an integer $<2k$, for which $g^{2l}=e$, which clearly contradicts the fact that order of $g$ is $2k$. Thus we see that $k$ is the smallest integer $n$ such that $(g^2)^n=e$. Hence if order of $g$ is $2k$, then order of $g^2$ is $k$.
If the orders of $g_1,g_2\in G$ be $a,b$ respectively, where $\mathrm{gcd}(a,b)=1$. Then if $G$ is abelian, order of $g_1g_2$ is $ab$. Obviously $(g_1g_2)^{ab}=g_1^{ab}g_2^{ab}$, as $G$ is abelian and $g_1^{ab}g_2^{ab}=(g_1^a)^b(g_2^b)^a=e$. If $r$ is the order of $g_1g_2$, then $r\mid ab$. Now $e=e^a=(g_1g_2)^{ra}=g_1^{ra}g_2^{ra}=g_2^{ra}$, as $g_1^{ra}=e$. This implies $b\mid ra\Rightarrow b\mid r$  as $b$ and $a$ are relatively prime. Similarly we can prove $a\mid r$, and as $a$ and $b$ are relatively prime we have $ab\mid r$. Thus we get $r=ab$.
Using these two results you can see that $m^2$ has order $4/2=2$ and $m^2n$ has order $2\times3=6$.
