# Prove - intersection of subgroups is a subgroup. [duplicate]

This question already has an answer here:

Since I have some similar question to this one, I'll be happy to understand how to solve it.

$G$ is a group and $A,B\le G$. I need to show that $A \cap B \le G$

(If $A, B,$ are subgroups of $G$, I need to show that $A \cap B$ is a subgroup of $G$, too.)

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## marked as duplicate by Grigory M, amWhy, Claude Leibovici, Sanath Devalapurkar, Sami Ben RomdhaneApr 26 '14 at 14:43

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What have you tried? This follows pretty immediately from definitions. –  JSchlather Jan 21 '13 at 21:48
Im not sure. is that mean that A and B are subgroups of G?? –  Mary Jan 21 '13 at 21:52
Yes, that is the usual meaning of $A\leq G$. –  Eric Stucky Jan 21 '13 at 21:54
so now that solve all my problems :) thanks –  Mary Jan 21 '13 at 21:59
So, the actual question was, "In group theory, what does the notation $A\le G$ mean?" –  Gerry Myerson Jan 21 '13 at 23:42

## 1 Answer

Clarification
In the context of group theory, if $G$ is a group, then $A \leq G$ denotes "$A$ is a subgroup of $G$"

$A$ and $B$ are subgroups of a group $G$ (i.e. $A\leq G, B\leq G$).
Prove that $A \cap B$ is a subgroup of $G$ (i.e, $A\cap B \leq G$).

Hint 1:
You know that $A$ and $B$ are subgroups of $G$. That means they each contains the identity element, say $e$ of $G$. So what can you conclude about $A\cap B$? If $e \in A$ and $e \in B$, then...? (Just unpack that means for their intersection.)

Hint 2:
You know that $A, B$ are subgroups of $G$. So they are both closed under the group operation of $G$. If $a, b \in A\cap B$, then $a, b \in A$ and $a, b\in B$. Since $A, B$ are subgroups of $G$, they are closed under the group operation of $G$, so we know it follows that $ab \in A$ and $ab \in B$. So what can you conclude about $ab$ with respect to its containment in $A\cap B$? This is about proving closure of $A\cap B$ under the group operation of $G$.

Hint 3:
You can use similar arguments to show that for any element $c \in A\cap B$, $c^{-1} \in A\cap B$:
If $c \in A\cap B$, then $c \in A$ and $c\in B$, so $c^{-1} \in A$ and $c^{-1} \in B$, again, since $A, B$ are subgroups of $G$. Hence, what can we say about the containment of $c^{-1}$ in $G$? That answer will establish that $A\cap B$ is closed under inverses.

Once you've completed each step above, what can you conclude about $A\cap B$ in $G$?

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