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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $a*b \in H$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.