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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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  • $\begingroup$ You don't need to check associativity of a subgroup - see here $\endgroup$ – TastyRomeo Feb 2 '17 at 22:57
  • $\begingroup$ You just have to prove it is stable by products and inverses – and non-empty, of course. $\endgroup$ – Bernard Feb 2 '17 at 22:59
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    $\begingroup$ 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. $\endgroup$ – pjs36 Feb 2 '17 at 23:02
  • $\begingroup$ 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... $\endgroup$ – user3724404 Feb 2 '17 at 23:06
  • $\begingroup$ your comment should be enough to prove its associative $\endgroup$ – asddf Feb 2 '17 at 23:08
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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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  • $\begingroup$ 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. $\endgroup$ – Marc Bogaerts Feb 3 '17 at 0:19
  • $\begingroup$ 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. $\endgroup$ – Namaste Feb 3 '17 at 0:25
  • $\begingroup$ @Marc Of course, you are free to post your comment as an answer, as unhelpful as it is in this situation. $\endgroup$ – Namaste Feb 3 '17 at 0:28
  • $\begingroup$ 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. $\endgroup$ – Marc Bogaerts Feb 3 '17 at 0:30
  • $\begingroup$ Indeed. You've got a good point, @Marc $\endgroup$ – Namaste Feb 3 '17 at 0:31

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