If $H$ is a nonempty finite subset of $G$ and $H$ is closed under multiplication, then $H$ is a subgroup of $G$ The line "If $H$ is a nonempty finite subset of $G$ and $H$ is closed under multiplication, then $H$ is a subgroup of $G$" appears as Lemma 2.4.2 in Topics in Algebra by Herstein.
Is this a correct statement? I don't understand the multiplication part - should it be "closed under the operation of the group" instead of multiplication?

The word "multiplication" is often synonymous with the phrase "group operation". For example, I might say:

We multiply $g$ and $h$ together to get $gh$.

Writing this general sentence using the term "group operation" is messy. I also might say:

The product of $g$ and $h$ is $gh$.

But here the word "product" is multiplication-specific, as for example $a+b$ is the "sum".
This usage is consistent with the use of the word "multiplication" in rings and things, as a ring is an abelian group with a multiplicative structure, and every group embeds into the multiplicative structure of some ring.

Note that "finiteness" is required in your result. If $H$ is an infinite set then there are counter-examples. In particular, set $G=(\mathbb{Q}, \times)$ and $H=\{x\in\mathbb{Q}\mid |x|\geq1\}$, then $H$ is closed under the group operation but is not a subgroup. As another example, take $G=(\mathbb{Z}, +)$ and $H=\mathbb{N}$ (note the use of additive notation, and that this doesn't matter).
