Finite group and subgroups 
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*$G$ 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.

*Can you help me find an example of a group $G$ that is not abelian, and such that $A =\{g\in G |g\neq g^{-1}\}\nleq G$.
 A: 
(1) Let $G$ be a finite group, with $A,B$ both subgroups of $G$, and $A\nsubseteq  B$. $\quad$Prove that $|A\cap B|$ $\le$ $(1/2)|A|$. 



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*You need only consider $|G| \geq 4$ if $A, B$ are subgroups of $G$, and $A \nsubseteq B$. (Otherwise, if A and B are both subgroups of $G$, then $A \subseteq B$.) 

*Be sure you know, or can prove, the fact that: If $A, B$ are subgroups of $G$, then $A\cap B$ is a subgroup of $G$.

*Yes, you can use Lagrange's Theorem.

(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 $G = S_3$.
A: For 1, can you prove that the intersection of two subgroups of $G$ is a subgroup of $G$?
A: 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. 
