**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.16**

Let $A$ and $B$ be groups. Prove that the maps $\pi_1 : A \times B \rightarrow A$ and $\pi_2 : A \times B \rightarrow B$ given by $\pi_1((a,b)) = a$ and $\pi_2((a,b)) = b$ are homomorphisms and find their kernels.

Solution: First to see that $\pi_1$ and $\pi_2$ are homomorphisms, let $(a_1,b_1), (a_2,b_2) \in A \times B$. Then $$\pi_1((a_1,b_1) \cdot (a_2,b_2)) = \pi_1((a_1a_2,b_1b_2)) = a_1a_2 = \pi_1((a_1,b_1)) \cdot \pi_1((a_2,b_2))$$ and $$\pi_2((a_1,b_1) \cdot (a_2,b_2)) = \pi_2((a_1a_2,b_1b_2)) = b_1b_2 = \pi_2((a_1,b_1)) \cdot \pi_2((a_2,b_2)).$$ Now we claim that $\mathsf{ker}\ \pi_1 = 1 \times B$.

($\subseteq$) Suppose $(a,b) \in \mathsf{ker}\ \pi_1$. Then $a = \pi_1(a,b) = 1$, so that $(a,b)\in 1 \times B$.

($\supseteq$) Suppose $(a,b) \in 1 \times B$. Then $\pi_1((a,b)) = a = 1$, so that $(a,b) \in \mathsf{ker}\ \pi_1$. The proof that $\mathsf{ker}\ \pi_2 = A \times 1$ is analogous.