Let $G$ be a group and $H_1,...,H_n$ subgroups. Consider the function $ \Phi:H_{1}\times...\times H_{n}\to G $
Defined by $$ \Phi\left(h_{1},...,h_{n}\right):=h_{1}\cdot...\cdot h_{n} $$
Prove that $\Phi$ is an isomorphism of groups, if and only if the following conditions holds:
a) $ H_{1}H_{2}...H_{n}=\left\{ h_{1}\cdot h_{2}\cdot....\cdot h_{n}:h_{i}\in H_{i},1\leq i\leq n\right\} =G $
b) $ H_{i} $ is a normal subgroup of $G$ for any $i$
c) $ \left(H_{1}\cdot....\cdot H_{i}\right)\cap H_{i+1}=\left\{ e\right\} $ for any i.
What I have already done:
I'm trying to prove by induction. I have already proved the base case - for $n=2$. Im not sure how to use the induction hypothesis properly in order to prove the induction step (At first I thought it would be intuitive and I'll just use the induction base for the $n=2$ case, but I'm having trouble with writing it in a formal way).
Any help would be appreciated. Thanks in advance.
First, let’s assume $\phi$ is a group isomorphism.
Condition (1) simply says that $\phi$ is a surjective function; since $\phi$ is an isomorphism, it’s a bijection and therefore a surjection. So that follows immediately.
Now suppose $ab \in H_i$. Write $a = \phi(a_1 \ldots a_n)$ and $b = \phi(b_1 \ldots b_n)$, where for all $i$, $a_i, b_i \in H_i$. Then $ab = \phi(a_1 \cdot b_1, \ldots, a_n \cdot b_n)$ since $\phi$ is a group homomorphism. Since $\phi$ is a bijection, we see that $a_j b_j = e$ whenever $j \neq i$ and that $a_i b_i = ab$. So $a_j = b_j^{-1}$ for all $j \neq i$. Therefore, $ba = \phi(b_1 a_1, \ldots, b_n a_n)$. Now for all $j \neq i$, we have $b_j a_j = e$. Therefore, $ba = b_i a_i \in H_i$. Then $H_i$ is normal. Condition 2 holds.
For the third condition, suppose we have $a_1 \cdot \ldots \cdot a_i = b$, where $a_j \in H_j$ for all $j$ and also $b \in H_{i + 1}$. Then we have $\phi(a_1, \ldots, a_i, b^{-1}, e, \ldots, e) = \phi(e, \ldots, e)$, and therefore, since $\phi$ is injective, $b = e$. Condition 3 holds.
So if $\phi$ is a group isomorphism, all the conditions hold.
Proving the converse does appear to require some induction.
In fact, to prove the converse, we prove a slightly different theorem.
Note that once we prove this, we will apply condition (1) to conclude that $\phi$ is an isomorphism.
Proof: we proceed by induction on $n$. The base case $n = 0$ is trivial, since the zero map $0 \to G$ is automatically an injective group homomorphism and $0$ is automatically normal.
Now suppose the statement holds for groups $H_1, \ldots, H_i$ with map $\phi_i$, and consider the map $\phi : H_1 \times \ldots \times H_{i + 1} \to G$. Let $A$ be the image of $\phi_i$. Then we can view $\phi$ as a map $\phi : A \times H_{i + 1} \to G$, where $A$ and $H_{i + 1}$ are normal with intersection $\{e\}$.
Note that given $a \in A$ and $h \in H_{i + 1}$, we see that $a$ and $h$ commute, since $a h a^{-1} h^{-1} \in A \cap H_{i + 1}$ and therefore $a h a^{-1} h^{-1} = e$, so $ah = ha$. Therefore, the map $\phi(a, h) = a \cdot h$ is a group homomorphism $A \times H_{i + 1} \to G$.
Furthermore, if $ah = a’ h’$, then $h h’^{-1} = a^{-1} a’$, and therefore both sides of the equation are in the group $A \cap H_{i + 1} = \{e\}$, so $h = h’$ and $a = a’$. Therefore, $\phi$ is injective.
Finally, suppose $a \in A$, $x \in G$, and $h \in H_{i + 1}$. We have $x a h x^{-1} = (x a x^{-1})(x h x^{-1})$, so the image of $\phi$ is indeed normal.