Show that a Kernel of a natural mapping in normal groups is the intersection of these.

Question

I've got the following exercise extracted from Rotman's Introduction to Theory of Groups: Let $H_1, \dots , H_n $ the normal sub groups of $G$, we define $\phi : G \rightarrow G/H_1 \times \dots \times G/H_n$ like $$\phi(x) = (xH_1, \dots , xH_n)$$ show that $ker(\phi) = H_1 \cap \dots \cap H_n$.

Well we know that the trivial group is part of the kernel and the intersection as well, but I cannot guarantee that is the whole intersection. So any hints?

Answer 1

There are two steps:

1) To show $Ker(\phi)\subseteq\cap_{i=1}^n H_i$

Let $x\in Ker(\phi)$, then $\phi(x)=(xH_1,\dots,xH_n)=(1H_1,\dots,1H_n)$ so $x\in H_i$ for each $i=1,\dots n$ so $Ker(\phi)\subseteq\cap_{i=1}^n H_i$.

2) To show $\cap_{i=1}^n H_i \subseteq Ker(\phi)$

Now let $y \in \cap_{i=1}^n H_i$, then $\phi(y)=(yH_1,\dots,yH_n)$ but $y\in H_i$ so $yH_i=1H_i$ thus $\phi(y)=(1H_1,\dots,1H_n)$ and we conclude $y\in Ker(\phi)$. In particular $\cap_{i=1}^n H_i \subseteq Ker(\phi)$.

So we can conclude $Ker(\phi)=\cap_{i=1}^n H_i$.