If $H_1,\dots, H_s$ are normal subgroups of $G$, show that the function:
$$\varphi: G\to G/H_1 \times\dots\times G/H_s$$ given by:
$$\varphi(g) = (gH_1,\dots, gH_s)$$
is a homomorphism with kernel:
$$\ker \varphi = \bigcap_{i=1}^{s}H_i.$$
I first started by proving that $\varphi$ is a group homomorphism.
Given $g,h$, let's check that $\varphi(gh)=\varphi(g)\varphi(h).$
Therefore as the group operation in $\varphi$ is preserved, $\varphi$ is a group homomorphism.
Do you think I'm doing them the right way? Because from here on I'm getting a little confused about what to do.
As for the kernel, note that: \begin{alignat}{1} \operatorname{ker}\varphi &= \{g\in G\mid \varphi(g)=(H_1,\dots,H_s)\} \\ &= \{g\in G\mid (gH_1,\dots,gH_s)=(H_1,\dots,H_s)\} \\ &= \{g\in G\mid (gH_1=H_1) \wedge\dots\wedge (gH_s=H_s)\} \\ &= \{g\in G\mid (g\in H_1) \wedge\dots\wedge (g\in H_s)\} \\ &= \{g\in G\mid g\in \bigcap_{i=1}^sH_i)\} \\ &=\bigcap_{i=1}^sH_i \end{alignat} Can you justify all the equal signs?