# Direct product of two normal subgroups

Let $A$ and $B$ be normal subgroups of a group $G$ such that $A \cap B = \langle e \rangle$ and $AB = G$. Prove that $A \times B \cong G$

Attempted proof:

Define $f : A \times B \rightarrow G$ by $f(a,b) = ab$. From a proof of another exercise, the hypothesis of this problem implies $ab = ba$. Therefore, $f(a,b) = f(b,a)$.

Since $f((a,b) \cdot (c,d)) = f(ac, bd) = (ac)(bd) = a(cb)d = a(bc)d = (ab)(cd) = f(ac) f(bd)$ for $a,c \in A$ and $b,d \in B$, it follows that $f$ is a homomorphism.

Now I need to show that $f$ is bijective (so show both injectivity and surjectivity of $f$), so that $f$ is an isomorphism from $A \times B$ to $G$.

Am I on the right track?

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Should be $f(ab, cd) = ab \cdot cd = f(a, b) \cdot f(c, d)$. Everything else sounds ok –  dani_s Feb 26 '14 at 8:33
Fixed the error. My bad. –  dragon Feb 26 '14 at 8:34
Ops I meant $f(a, c) \cdot f(b, d)$. I guess you are assuming $a, b \in A$ and $c, d \in B$ –  dani_s Feb 26 '14 at 8:36
Yes, that too ($a,b \in A$ and $c,d \in B$). –  dragon Feb 26 '14 at 8:39
What do you mean by "the hypothesis of this problem implies $ab=ba$"? The group certainly does not have to be abelian. Also $f(a,b)=f(b,a)$ doesn't mean anything, since $f$ is defined on $A\times B$, not on $B\times A$. And again, $f(ac)$ is even worse, since $f$ is a map of two elements, not one. –  Christoph Feb 26 '14 at 8:41

Hints: You will need $AB=G$ to prove surjectivity and $A\cap B=\{e\}$ to prove injectivity.

Since you still didn't fix the issue with $f(b,a)$ and $f(ac)$ being meaningless expressions, let me elaborate on this:

You want a homomorphism $f:A\times B\to G$, so you can put a pair $(a,b)\in A\times B$ in $f$ an get some element $f(a,b)\in G$.

• Writing $f(b,a)$ for $b\in B, a\in A$ implies you could put a pair $(b, a)\in(B\times A)$ in $f$, which is not the case, since $f:A\times B \to G$, $\color{red}{\text{not }f:B\times A\to G}$.
• Writing $f(ac)$ for $a,c\in A$, thus $ac\in A$, implies you could just put an element of $A$ in $f$, which is not the case, since $f:A\times B\to G$, $\color{red}{\text{not }f:A\to G}$.

What you want is for $(a,b), (c,d) \in A\times B$ to have $$f((a,b)\cdot(c,d)) = f(a,b)\cdot f(c,d),$$ where by definition of the operation in $A\times B$ we have $(a,b)\cdot(c,d)=(ac,bd)\in A\times B$. Now you suggest to define $f$ as \begin{align} f : A\times B &\longrightarrow G, \\ (a,b) &\longmapsto ab. \end{align} Using this definition, we can evaluate both sides of the equation we want to prove for $f$ to be a homomorphism: \begin{align} f((a,b)\cdot(c,d)) &= f(ac, bd) = acbd, \\ f(a,b)\cdot f(c,d) &= ab\cdot cd = abcd. \end{align} Now, since $c\in A$, $b\in B$ where $A,B$ are normal in $G$ and $A\cap B=\{e\}$ you conclude from your previous problem that $$acbd = a(cb)d = a(bc)d = abcd.$$ Put everything together and you obtain \begin{align} f((a,b)\cdot(c,d)) &= f(ac, bd) = acbd = a(cb)d \\&= a(bc)d = abcd = ab\cdot cd = f(a,b)\cdot f(c,d). \end{align}

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he didn't ask for hints... –  dani_s Feb 26 '14 at 8:50
@dani_s I'm still working this problem out on paper, so actually, hints like this one are helpful. :) –  dragon Feb 26 '14 at 8:53
@glacier still, if you didn't ask for hints then no hints should be given! –  dani_s Feb 26 '14 at 8:56
What you have to prove is that if $f(a, b) = f(c, d)$ then $(a, b) = (c, d)$ –  dani_s Feb 26 '14 at 9:15
@glacier No, because $a \in A$ and $b \in B$ doesn't imply $ab \in A \cap B$ –  dani_s Feb 26 '14 at 9:38

Perhaps more succint and going the other way around (easier, imo):

$$G=AB\implies \forall\,x\in G\;\;\exists\,a_x\in A\,,\,b_x\in B\;\;s.t.\;\;x=a_xb_x$$

Now, if also $\;x=a'_x+b'_x\;,\;\;a'_x\in A\;,\;\;b'_x\in B\;$ , then we'd have that

$$a_xb_x=a'_xb'_x\implies a'^{-1}_xa_x=b'_xb_x^{-1}\in A\cap B=1\implies\;\ldots$$

and we get injectivity of $\;G\to A\times B\;$ defined by $\;x\mapsto (a_x,b_x)\;$

Surjectivity is trivial, of course.

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