$\mathbb{K}^{\times} \times \mathbb{Z}$ is not group isomorphic to $\mathbb{K}^{\times}$ Let $\mathbb{K}$ be a field with infinite cardinality and $\mathbb{K}^\times$ its group of units under multiplication (i.e. all elements except $0$). I want to determine if $\mathbb{K}^{\times} \times \mathbb{Z}$ is group isomorphic to $\mathbb{K}^{\times}$ or not. Intuitively I'm inclined to say not, since the first "has a lot more elements" than the second, but I'm not sure because weird things can happen with infinite groups, like proper subgroups being isomorphic to the whole group and so on.
Just to be clear, the operation equipped on $\mathbb{K}^{\times} \times \mathbb{Z}$ is $(k,m) \star (h,n) = (k\cdot h, m+n)$.
This question was inspired by this post. In the second answer, this same question is solved for $\mathbb{K} = \mathbb{C}$, using the fact that $\mathbb{C}^\times$ is a divisible group. But that's not true for $\mathbb{R}^\times$ for example, so I cannot use it in the general case. Can anyone help me?
 A: $\newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q}$ Actually I think the claim to be false. A counter example should be given by $\Q^*$.
The short version is to observe that $\Q^*$ is isomorphic to $\Z/2\Z \times \bigoplus_{n \in \N} \Z$, then by the following chain of isomorphisms
$$\Z/2\Z \times \bigoplus_{n \in N} \Z \cong \Z/2\Z \times \Z \times \bigoplus_{n \in N} \Z \cong \Z \times \Z/2\Z \times \bigoplus_{n \in N} \Z \cong \Z \times \Q^*$$
it follows that $\Q^* \cong \Z \times \Q^*$.
To see the isomorphism between $\Q^*$ and $\Z/2\Z \times \Z$ consider the following map
$$
\begin{align*}
f \colon \Z/2\Z \times \bigoplus_{n \in \N} \Z & \longrightarrow Q^* \\
f(s, \bar a) = (-1)^s \prod_{n \in \N} p_n^{a_n}
\end{align*}
$$
where $s \in \Z/2\Z$, $\bar a =(a_n) \in \bigoplus_n \Z$ and for each natural number $n$ the number $p_n$ is the $n$-th prime number in $\N$.
By a simple calculation one can prove that $f$ is indeed a group homomorphism:
$$
\begin{align*}
f((s,\bar a)+(t,\bar b)) &= f((s+t,\bar a + \bar n)) \\
&= (-1)^{s+t} \prod_n p_n^{a_n+b_n} \\
&= (-1)^s \prod_n p_n^{a_n} (-1)^t \prod_n p_n^{b_n} \\
&= f((s,\bar a)) f((t,\bar b))
\end{align*}
$$
The injectivity follows by observing that $f((s,\bar a)) = 1$ if and only if
$(-1)^s\prod_n p_n^{a_n} = 1$, which happens only if $s=0$ and for every $n$ we have $p_n^{a_n}=1$, that is if $a_n=0$.
Surjectivity follows observing that the image of $f$ contains all the integers which generate $\Q^*$.
I hope this helps.
