Will $G$ have an element $ab$ of order lcm$(|a|, |b|)$ if $|a|$ and $|b|$ are not necessarily relatively prime? 
If $G$ is a finite abelian  group, $a, b\in G$ such that $|a|=m>1, |b|=n>1$ where $m, n$ are not necessarily relatively prime, then prove or disprove that $G$ has an element $ab$ of order lcm$(m, n).$

My claim: It is not true in general. For example consider $G=\langle x\rangle $ with $|x|=4, a=x$ and $b=x^3$. Then we can see that $|a|=4=|b|$ but $|ab|=1\neq lcm(4, 4)$.
Am I correct in my logic? Please help me. Thanks in advance
 A: You are correct that the proposition is false, if it's a proposition that is being claimed for all products $ab$ of arbitrary $a\,b, \in G$. And your counterexample is a very good one: provided you specify that $|G| = |\langle x\rangle| = 4$ (so you need to make that explicit).  
For the sake of transparency, perhaps make explicit what you want to denote as $a$, as $b$, and $G$. For example, we could put $G = \mathbb Z_4$ (the additive group of integers modulo $4$, with specific elements $a = 1$, $b = 3$ such that $|a| = |b| = 4$.
Then as you rightly claim, $|a| = m = 4 = n = |b|$, but $|ab| = 1 \neq \operatorname{lcm}(m, n) = 4$.

Unless $\gcd(|a|, |b|)=1$ is given we can't say that order of $ab$ is same as lcm$(|a|, |b|)$. We can only make such a claim provided $G$ is abelian and $\gcd(|a|, |b|) = 1$, i.e., if the orders of $a$ and $b$ are relatively prime.

ADDED:
Given the OP's edited question: Existence is easily established.
In every group $G$, the following holds: $\exists e \in G$, so put $a = b  = e$. Then $ab = e^2 = e$ and $|ab| = |e| = 1 = \operatorname{lcm}(|a|, |b|)$. 
So trivially, yes, for every group, there exists $(a, b)$ such that $|ab| = \operatorname{lcm}(|a|, |b|)$.
A: Let $G$ be a finite abelian group. Then if $a,b$ such that $ab \neq e_G \in G$ with $|a|=m$ and $|b|=n$, we know that the order of $|ab|=\text{lcm}(m,n)$ (or is a multiple of the order). But since $G$ is closed under its operation, we know that $ab \in G$. But since $ab \in G$ and $|ab|=\text{lcm}(m,n)$, the result holds (if it is a multiple, we can still construct such an element, I encourage you to see why). It is for you show that that in any abelian group that $|ab|=r \mid \text{lcm}(|a|,|b|)$, their orders not necessarily being finite. 
EDIT. Note that I am only stating the statement is true if $ab \neq e_G \in G$. If we allow this, then indeed the result is false. Otherwise, the group you are suggesting, $\mathbb{Z}_4$, the integers under addition modulo $4$, with $a=1$ and $b=3$ works as counterexample (note that $1+3=4\equiv 0 \mod 4$ so that $ab=0=e_G \in G$).
A: Thanks to all of you for giving such helpful response. Your valuable discussion has helped me a lot. Based on whatever the information I received, here is my solution.
Let $a, b\in G$ with $|a|=m>1, |b|=n>1$. Assume that $|ab|=\alpha$. Let $\gcd(m, n)=d$ and $lcm(m, n)=L$. Then $mn=dL$ i.e. $\frac{mn}{d}=L$.
Now $gcd(m, n)=d\Rightarrow gcd(\dfrac m d, \dfrac n d)=1$. 
Consider the element $a^d b^d$. Then $(a^d b^d)^{\frac{mn}{d}}=(a^d)^{\frac{mn}{d}} (b^d)^{\frac{mn}{d}}$ because we know that in an abelian group $G$, $(ab)^t=a^t b^t$ for every integer $t$.
From the orders of $a$ and $b$, so it follows that $(a^d b^d)^{\frac{mn}{d}}=e$ where $e$ is identity element in $G$.
Thus $\alpha | \frac{mn}{d}=L$ i.e. $\alpha|L$. Once we prove that $L|\alpha$ we are done.
So let $L\nmid \alpha$. Note that here now we have $L<\alpha$. By division algorithm we then write $\alpha = Lq+r, 0\leq r < L$. If $r=0$ then $L|\alpha$. So let $0<r<L$.
then $(a^d b^d)^r=(a^d b^d)^{\alpha-Lq}=(ab)^{d(\alpha-Lq)}=e$ (WHY?) which is contradiction as $r<L<\alpha$ where $\alpha=|ab|$. Thus we must conclude $L|\alpha$.
Hence $L=\alpha$ viz $|a^d b^d|= lcm(m, n)$.
Hence the proof.
Am I correct?
