If $ab$ is an element of group $G$, are $a$ and $b$ both elements of group $G$ as well? Obviously, if $a$ and $b$ are elements of group $G$, then $ab$ is in $G$ as well.
Is the converse true?
I tried to think about it in terms of the inverse of the original statement (considering it's the contrapositive of the statement I'm trying to prove), but I wasn't sure how to prove that that was true (i.e. if one or both of the elements are not in $G$, then $ab$ is not in $G$).
 A: Consider the groups $\Bbb Z$ and $2\Bbb Z$.  $1,3\notin 2\Bbb Z$, but $1+3\in 2\Bbb Z$.
A: We can consider the group $(\mathbb{Z}^*,\times)$ $\sqrt2 \times \sqrt2 =2 $ while $\sqrt 2 \notin \mathbb{Z}^*$ Hence, the converse is not true.
A: The converse is not true: Think about the group of integers, $\mathbb{Z}$, (under addition) included in the group of rational numbers, $\mathbb{Q}$, (under addition). Then $\frac{1}{2}+\frac{1}{2}=1\in\mathbb{Z}$, but $\frac{1}{2}\notin\mathbb{Z}$.
A: By writing $ab$, you are assuming that it makes sense to concatenate the symbols a and b in a meaningful way. So there are some basic assumptions being made behind the scenes.
I will reinterpret your question like this. Suppose $a,b$ are in a group $G$ and let $H$ be a subgroup of $G$. If $ab \in H$, then are $a$ or $b$ in $H$?
This is False. For instance, within the reals under addition, we see that $1 + \sqrt 2$ and $1 - \sqrt 2$ have sum within the rationals, but neither are rational.
A: Let $H$ be a subgroup of a group $G$ with the property that whenever $ab \in H$ for some $a,b \in G$, then $a \in H$ or $b \in H$. Then $H=G$. For suppose $H \subsetneq G$. Let $a \in G-H$. And put $b=a^{-1}$. Then $ab=1 \in H$, since $H$ is a subgroup. But then by the property, $a$ or $a^{-1}$ are elements of $H$, whence $a \in H$, contradicting the choice of $a$.
