# The intersection of an abelian normal subgroup $A$ and subgroup $B$ is normal in $AB$

Let $G$ be a finite group with two subgroups $A$ and $B$. Show that if $A$ is Abelian and is normal in $G$, then $A\cap B$ is normal in $AB.$

It's been awhile since I did Abstract Algebra, (this is a prelim question) the above seems easy but still I want to check my proof:

We will first show that $A\cap B\leq AB.$ Observe that $A\cap B$ is a subset of $AB$ since if $x\in A\cap B$ then $x\in A$ and $x\in B$ and thus( for $x\in A$ and $e\in B$) $x\in AB=\{ab: a\in A \text{ and } b\in B\}.$ Clearly, $e \in A\cap B.$ Suppose that $x,y\in A\cap B.$ Then $x,y\in A$ and $x,y\in B.$ Since $A,B\leq G$, we have that $xy\in A$ and $xy\in B$, and thus $xy\in A\cap B.$ Finally, if $x\in A\cap B$ then $x^{-1}\in A\cap B$ since $x,x^{-1}\in A$ and $x,x^{-1}\in B.$ Now we will show that $A\cap B \triangleleft AB.$ Observe that if $k\in A\cap B$ and $g\in AB$ then it suffices to show that $gkg^{-1}\in A\cap B.$ Now consider, $\forall k\in A\cap B$ and $g\in AB$: \begin{align*} gkg^{-1}&=(ab)k(ab)^{-1} &&\text{ (where }a\in A \text{ and } b\in B)\\ &=a(bkb^{-1})a^{-1}\\ &=ak'a^{-1}&&\text{ (Since } A\triangleleft G \, , k'=bkb^{-1}\in A)\\ &=k'aa^{-1}&&\text{ (Since } A \text{ is Abelian )}\\ &=k'e\\ &=k'\in A\\ &\implies gkg^{-1}=k'\in A\cap B\\ \end{align*}

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Please make titles informative as to the content of the post, not your intentions. Asking people to check your proof is properly done in the body of the post, not in the title (which was completely uninformative as to what the question was about). –  Arturo Magidin Jun 26 '12 at 18:50
@ArturoMagidin I will keep that in mind next time. Thanks. –  Lyapunov Jun 26 '12 at 19:50

The first part is too complicated: $A\subseteq AB$ (whether or not $AB$ is a group), and $A\cap B\subseteq A$; so you are done as far as inclusion; and intersection of subgroups is always a subgroup, so $A\cap B$ is a subgroup and is contained in $AB$. Of course, you should note that $AB$ actually is a subgroup, because $AB=BA$ by the normality of $A$.

Your final assertion does not follow. You have justified correctly that $k'\in A$; but you have not said a single word as to why it is in $B$. So how do you justify your assertion that $k'\in A\cap B$?

The reason it is in $B$ is that $k\in A\cap B$, so $k\in B$. Therefore, $bkb^{-1}$, being a product of elements in $B$, must be in $B$. Since it is also in $A$ (by the reasons given), you h ave that $k'=bkb^{-1}\in A\cap B$.

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You are right about justifying $k'\in B$. I was thinking of union even though I wrote intersection! –  Lyapunov Jun 26 '12 at 19:54

Here is another solution: I don't mean that it's better, but I think you might pick up some tricks from it.

When you want to prove normality of a subgroup, it's sometimes useful (and certainly logical) to look at $N_G(H):=\{g\in G\mid gHg^{-1}\subseteq H \}$. (This is called the normalizer of $H$ in $G$.

1. Lemma: $N_G(H)$ is a subgroup of $G$, and in fact it is the largest subgroup of $G$ in which $H$ is normal.

2. Lemma: The product of a subgroup of $G$ and a normal subgroup of $G$ is a subgroup of $G$. Thus, $AB$ is a subgroup of $G$.

3. Since $A\lhd G$, then certainly $A\lhd AB$. By an isomorphism theorem$^\dagger$, $A\cap B\lhd B$.

4. $A\cap B$ is certainly normal in $A$ (and so is every other subgroup of $A$: Why?)

5. Observe $A$ and $B$ are in $N_G(A\cap B)$, hence the subgroup $AB\subseteq N_G(A\cap B)$. Done! (by the first lemma!)

$\dagger$ The one I am thinking of says if $A\lhd G$ and $B<G$, then $A\cap B\lhd B$ and $AB/A\cong B/(A\cap B)$.

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