show H is a subgroup Let G be a finite abelian group. Let $H=\langle a,b\rangle = \{a^{i}b^{j}\;\;\;i,j\in\mathbb{Z}\}$
Show that $H$ is a subgroup of $G$
My solution.
1) Need to show $H\neq\emptyset$
This is true as we can take $i=j=0$ then $a^{0}b^{0}=1\in H$
2) closure 
let $a^{i}b^{j}\in H$ and let $a^{k}b^{r}\in H$ then $a^{i}b^{j}a^{k}b^{r}=a^{i+k}b^{j+r}\in H$
3) let $a^{i}b^{j}\in H$ then $a^{-i}b^{-j}\in H$ as we can take $i=j=-1\in\mathbb{Z}$
So $H$ is a subgroup of G
Is this correct?
Thank you
 A: In 3), if you take $i=j=-1$, you have $a^ib^j=a^{-1}b^{-1}$. Better say: since $i,j\in \mathbb{Z}$, then also $-i,-j$ are integers, thus $a^{-i}b^{-j}$ is in $H$ as well.
A: Here's a pretty little factoid which you all may find intrigueing:  (2) is sufficient to establish that $H$ is a subgroup of $G$; this follows from the finiteness of $G$.  Indeed we have that if $S$ is a finite, closed (under the group operation) subset of any group $G$, then $S$ is a subgroup of $G$.  For if $s \in S$ we have $s^n$ in $S$ for all positive integers $n$, by closure.  Finiteness of $S$ implies the sequence $s^n$ must repeat itself; thus $s^i = s^j$ for some positive integers $i < j$, whence $1 = s^{j - i} \in S$.  Also, since $ss^{j - i -1} = s^{j - i} =1$, we have $s^{-1} = s^{j - i -1} \in S$.  Thus $S$ is a subgroup of $G$.  In the case at hand, $G$ finite implies $H$ finite and the desired result follows.  Essentially, $\text{(2)} \Rightarrow \text{(1) and (3)}$ in the case of $H$ finite.QED
Nota Bene:  This technique may be used to extend the result to finite but non-abelian $G$.End of note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
