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  1. $G$ is finite group. $A,B$ are subgroup of $G$. $A \nsubseteq B$. I need to prove that $|A\cap B|$ $\le$ $\frac{|A|}{2}$.

    I think it might have something to do with Lagrange but I can't find how it helps me.

  2. Can you help me find an example of a group $G$ that is not abelian, and such that $A =\{g\in G |g\neq g^{-1}\}\nleq G$.

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What does $A/2$ mean? –  MJD Jan 21 '13 at 22:44
    
|A|*1/2 ill edit my question –  baaa12 Jan 21 '13 at 22:45
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Do you mean $|A \cap B|$ divides $(1/2)|A|$? –  amWhy Jan 21 '13 at 22:51
    
no ... look no at th question –  baaa12 Jan 21 '13 at 22:52
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2 makes no sense –  user641 Jan 21 '13 at 22:53
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3 Answers 3

up vote 4 down vote accepted

(1) Let $G$ be a finite group, with $A,B$ both subgroups of $G$, and $A\nsubseteq B$.
$\quad$Prove that $|A\cap B|$ $\le$ $(1/2)|A|$.

(2). Given a non-abelian group G, such that such that if $A = \{g \in G: g \neq g^{-1}\}$, then prove $A$ is not a subgroup of $G$.

In this case, since if $e$ is the identity of $G$, then $e \in G, e^{-1} = e \in G$, so $e \notin A$, hence $A$ cannot be a subgroup of $G$, because it fails to contain the identity of $G$.

This works for any group $G$: nonabelian or otherwise.

If you need a specific example for a non-abelian group, pick, say, $G = S_3$: the non-abelian group of all permutations of the set $S = \{1, 2, 3\}$. $\;S_3 = \{(1), (1\,2), (1\, 3), (2\, 3), (1 \,2\, 3), (1\,3\,2)\},\;$ where $\,(1)\,$ is the identity permutation of $S_3 = G$.

Then let $A = \{(1\, 2\, 3), (1\, 3\, 2)\}\subset G$. That is, $A$ is the set of all elements $\,g \in G = X_3$ such that $g \in G$, $g \neq g^{-1}$. [Note, for each of the other elements of $G = S_3$: $(1) = (1)^{-1} \notin A$, $(1 \, 2) = (1 \, 2)^{-1} \notin A$, $(1\,3) = (1\, 3)^{-1} \notin A$, and $(2 \, 3) = (2\,3)^{-1} \notin A$.]

The fact that $\,(1) \notin A\,$ means that $\,A\,$ does not contain the identity element of $\,G = S_3,\,$ and therefore, $A$ fails to be a subgroup of $G = S_3$.

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I need an example to such a group not to prove that:/ but thanks! sorry I guess I go you wrong in my comment –  baaa12 Jan 21 '13 at 23:19
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Please edit the body of your question so it asks what you want to ask. Don't bury corrections in comment threads. –  Gerry Myerson Jan 21 '13 at 23:21
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And please, when you edit, indicate so, below your posted question, so that answers do not appear completely irrelevant. –  amWhy Jan 21 '13 at 23:22
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I wonder whether OP really wants an example of a nonabelian group where the $g$ with $g=g^{-1}$ don't form a subgroup. –  Gerry Myerson Jan 21 '13 at 23:27
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baaa12 I can work on the reversed question, baaa12, but perhaps post that as a separate question, as I put in a lot of time to answering the question above, as stated! At least give me credit for that! (and please, reaccept?) But do post your question as a new question (tmore he second part, with the equality stated). It will get more attention that way. –  amWhy Jan 22 '13 at 18:20
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For 1, can you prove that the intersection of two subgroups of $G$ is a subgroup of $G$?

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For the first question: Since things are finite, Lagrange's theorem tells us that the cardinality of a subgroup has to divide the cardinality of the big group. From this we know that a proper subgroup of a group has cardinality at most one half of the cardinality of the big group. Since $A$ is not contained in $B$, then the intersection of them must be a proper subgroup of $A$ and you are done.

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It is not clear that OP knows that the intersection of two subgroups is a subgroup (hence, my answer). –  Gerry Myerson Jan 21 '13 at 23:25
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