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To me this just reads: We suppose this and that and because we supposed it, it must be true.

Why does this show that $H$ is closed under inverse? Why does it show that $H$ is closed under the operation? It doesn't, it just assumes it.

Am I misunderstanding this?

user3200098
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  • Whether it is a good proof or not, it is a proof. Just read it. – Thomas Feb 18 '14 at 20:31
  • I have read it about 30 times and written it about 5 times on a blackboard while scribbling it about 10 times in my notebook with various thoughts and angles on it. It makes no sense to me. – user3200098 Feb 18 '14 at 20:32
  • It is a valid proof. Be more specific, whoat do you not understand? – Yiorgos S. Smyrlis Feb 18 '14 at 20:33
  • What's Theorem 3.1? – Malice Vidrine Feb 18 '14 at 20:34
  • That H is closed under multiplication and taking inverses are both premises of the theorem. – Vincent Pfenninger Feb 18 '14 at 20:35
  • Well can you tell me what he is looking for? Is he trying to show that $ab^{-1}$ exists in H? And if it exists in H by assuming that it is closed under operation and inverse, then it must be a subgroup. Why is it a subgroup if it is closed under operation and inverse? – user3200098 Feb 18 '14 at 20:35
  • It would be useful to know what theorem 3.1 is and what definition of a subgroup this book uses? – Vincent Pfenninger Feb 18 '14 at 20:35
  • Theorem 3.1: http://imgur.com/5SDySEu – user3200098 Feb 18 '14 at 20:36
  • I must admit that I find this proof (of the second theorem) weird as well. What is the definition of subgroup the book uses? Going by the proof of the first theorem, it is something like "$e \in H$; $H$ is closed under the operation; $H$ is closed under taking inverses." But, except for "$e \in H$", those are exactly the assumptions of the second theorem; and appealing to the first theorem just to get $e \in H$ seems overkill as the same argument that is used in the first theorem works just as well. – Magdiragdag Feb 18 '14 at 21:26
  • There are expositions of group theory where the fact that $ab^{-1}$ is in a subset $S$ whenever $a$ and $b$ are members is used to characterise a subgroup. From $a\in S$ and $a\in S$ we deduce that $1=aa^{-1}\in S$. Then from $1\in S$ and $a\in S$ we deduce $a^{-1}=1a^{-1}\in S$. Then from $a\in S$ and $b\in S$ we deduce $b^{-1}\in S$ whence $ab=a(b^{-1})^{-1}\in S$. So the one condition suffices. But in this case most of what we use the condition to prove is given already (hence some confusion). However the proof may simply be intended to illustrate the condition. – Mark Bennet Feb 18 '14 at 22:24

2 Answers2

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Read carefully for the hypotheses and the conclusion:

Hypotheses: $H$ is a subset of $G$, $H$ is closed under the operation, $H$ is closed under inverses.

Conclusion: Then $H$ is a subgroup of $G$.

Normally, given a random subset $H$ of $G$, we would have to check that it is closed under the operation, that it is closed under inverses, and that the identity is in $H$. (or maybe your book has a slightly different definition of subgroup, but the main point is, the definition requires you to check a (longish) list of properties)

This theorem is saying, in essence, that if the first two properties are satisfied, the third will be automatically satisfied (i.e. there is no need to verify that whole long list from the definition). The referred-to theorem (3.1) reduces this to showing $ab^{-1} \in H$. (i.e. {closed under inverses} and {closed under operation} $\Longrightarrow$ $ab^{-1} \in H \Longrightarrow H$ is a subgroup)

Explanation of Theorem 3.1:

Here, we want to show that satisfying the hypothesis "$ab^{-1} \in H$ whenever $a,b\in H$" is enough to conclude $H$ is a subgroup. That is, in the future, if we can simply check $ab^{-1} \in H$, there is no need to verify the usual requirements of a subgroup. (looks like this list is "operation is associative", "closed under inverses", "closed under operation" and "identity is in the set")

Assuming you understand how the other properties are verified, we need to show the identity is in such a set. If we choose $a = x$, $b = x$, then we know $ab^{-1}$ is in $H$. But $ab^{-1}$ is nothing more than $xx^{-1} = e$, so that is verified.

Many of these beginning abstract algebra proofs will seem... tautological sometimes. (i.e. you may find yourself thinking "what is there to even prove?!?") Take it as an exercise in patience and careful understand of definitions (and how to verify them), and take comfort in the fact that things will become a lot less trivial in time.

BaronVT
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It seems to me that the proof assumes that associativity is inherited (it doesn't prove that the operation on $H$ is associative, but it is associative in $H$ because associative in $G$), and all you really have to do to prove that $H$ is a subgroup is to show it contains the identity - because then it is closed under an associative binary operation with identity and inverses.

For this we need the condition that $H$ is nonempty, which isn't used explicitly in the proof as stated. Take $a\in H$ (which exists because $H$ is nonempty) then by the existence of inverses $a^{-1}\in H$ and then by closure under the binary operation $aa^{-1}=1\in H$.

Mark Bennet
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