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In order to prove something is a subgroup of $G$, you must prove it is a group - one of which criteria is that it is associative. Any tips for proving associativity in these situations? I'm thinking saying something like:

Fix arbitrary $a,b,c$ are in $H$. Then they must be in $G$ since they are in $H$. Since $G$ is associative, $(ab)c = a(bc)$. So $H$ is associative.

Is this sufficient? By the way, this is for an undergrad Modern Algebra course.

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    You don't need to check associativity of a subgroup - see [here](http://math.stackexchange.com/questions/1022907/why-dont-we-need-to-check-for-associativity-for-a-subgroup) – 2017-02-02
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    You just have to prove it is stable by products and inverses – and non-empty, of course. – 2017-02-02
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    Before we rush to say that certain things aren't necessary to determine if one has a subgroup of a given group, consider that OP might not have seen any subgroup tests yet -- so they really might have to check that subgroups are, among other things, groups. – 2017-02-02
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    I don't know what a subgroup test is :) I should also mention that subgroups were only introduced this week. Do you think my instructor would be satisfied by saying "H's associativity follows from G's"? I suppose I may as well ask... – 2017-02-02
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    your comment should be enough to prove its associative – 2017-02-02

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Let $H \subseteq G$, where $G$ is a group under the operation *.

Yes, in short, your argument is sufficient to conclude the group operation on G, and hence on $H\subseteq G$, is associative:

Since $H$ is a subset of $G$, every element $a,b, c \in H$ is an element of the group $G.$ Since G is a group by hypothesis, associativity of the group operation on $G$ holds also for $H$ because $H\subseteq G$.

To prove $H$ is a subgroup of $G$, of course, you must also show the identity element of $G$ is in $H$. And, you must also show that the inverse of any $a\in H$ is also in $H$.

Depending on what you've learned about groups, subgroups, etc., it never hurts, also, until you learn more concise tests for groups/subgroups, to ensure that $H\subset G$ has closure under the group operation. That is, for any $a, b\in H$, we must have $a*b \in H$.

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    For a subset $H$ of $G$ to be a subgroup it suffices to show that $\forall a,b \in H$ we have $ab^{-1} \in H$ this covers all the possibilities: to show the existence of the identity one takes $b = a$. Then since $e \in H$ one can take $a=e$ which proves the existence of the inverse. Since the inverse exists one can take $b=b^{-1}$ which proves closure for the group operation. – 2017-02-03
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    Yes indeed, @Marc. Please note my statement in my answer: "Depending on what you've learned about groups, subgroups, etc., it never hurts, also, **until you learn more concise tests for groups/subgroups,...**. Clearly, I gear my answer here to an undergraduate student in their first algebra course (two weeks into the semester?). I intentionally left out the "**more concise test**" you refer to. – 2017-02-03
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    @Marc Of course, you are free to post your comment as an answer, as unhelpful as it is *in this situation.* – 2017-02-03
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    Ok, I agree, but other people also read these comments and it's a nice trick to know, that"s why I left it as a comment and not as an answer. – 2017-02-03
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    Indeed. You've got a good point, @Marc – 2017-02-03