Exercise A0.II.3.4

Proposition

There exists a group $G$ such that $G \cong G \times H$ for a non-trivial $H$.

Proof

Let $G$ be the (abelian) group of infinite binary sequences

$$[v_0, v_1, v_2, \dots], v_n \in \{0, 1\}$$

under componentwise addition modulo $2$ (notated as $\oplus$). We will demonstrate that $G \cong G \times G$.

Let $a, b \in G$ and let $f : G \times G \to G$ be the interleaving map

$$f : ([v_0, v_1, \dots], [u_0, u_1, \dots]) \mapsto [v_0, u_0, v_1, u_1, \dots].$$

We can recover $a$ and $b$ from $c = f(a,b)$ since $a = [c_0, c_2, \dots]$ and $b = [c_1, c_3, \dots]$. This demonstrates that $f$ is a bijection. Next, we will show that $f$ is a homomorphism.

Let $a, b, a', b' \in G$ and $z = f\bigl((a,a') \bullet_{G \times G} (b, b')\bigr)$, where $\bullet_{G \times G}$ is the group operation of $G \times G$. Using the definition of direct product, we can express $z$ as $f(a \oplus b, a' \oplus b')$, which means that

$$\begin{align*} z &= [a_0 \oplus b_0, a'_0 \oplus b'_0, a_1 \oplus b_1, \dots] \\ &= [a_0,a'_0,a_1,\dots]\oplus[b_0,b'_0,b_1,\dots] \\ &= f(a,a')\oplus f(b,b'). \end{align*}$$

We conclude that $f$ is a homomorphism. Since it’s a homomorphism and a bijection, it is an isomorphism.

$\blacksquare$