Infinite groups such that $G/G'$ has odd order. Can someone give examples of an infinite group $G$ such that $G/G'$ has odd order.
 A: Take the direct product of an infinite simple group (or infinite direct product of finite simple groups) with a cyclic group of order $p$ , an odd prime.
A: Free products
A different style of answer would be to take the free product $G$ of two groups $H$ and $K$ of odd order, so $G=H\ast K$. Then the abelianisation $G/G^{\prime}$ is $H\times K$ which has odd order. For example, take $H$ to be cyclic of order $3$ and $K$ to be cyclic of order $5$.
$$G=C_3\ast C_5\cong\langle a, b; a^3, b^5\rangle$$
Then $G$ has the following abelinisation.
$$\langle a, b; a^3, b^5, [a, b]\rangle$$
This is $C_3\times C_5$, which is cyclic of order $15$.
It is not clear that the group $G$ in my example above is actually an infinite group, unless you know something about free products. To see that it is infinite, you can investigate its action on the tree where vertices have degree $3$ or $5$, and no two adjacent vertices have the same degree. For more details on this idea, see this post of mine, from the $5^{th}$ paragraph from the top (I link to this not just because of the example, but because hopefully the rest of the post will convince you that this viewpoint is useful). For more general details on "free products", look up the books of Magnus, Karrass and Solitar and of Lyndon and Schupp; both books are entitled Combinatorial group theory, and are the classic texts in combinatorial group theory.
A torsion-free example
It would be nice if we could make $G$ torsion-free, as then be aren't simply piggy-backing on a finite group. The following group works (it is "borrowed" from something called Rips' construction). However, I am sure that a more elementary construction exists!
$$
\begin{align*}
G=\langle
a, b, x;\\
x^3&=abab^2ab^3ab^4\cdots ab^i\\
xax^{-1}&=ab^{i+1}ab^{i+2}ab^{i+3}ab^{i+4}\cdots ab^{j}\\
x^{-1}ax&=ab^{j+1}ab^{j+2}ab^{j+3}ab^{j+4}\cdots ab^{k}\\
xax^{-1}&=ab^{k+1}ab^{k+2}ab^{k+3}ab^{k+4}\cdots ab^{l}\\
x^{-1}ax&=ab^{l+1}ab^{l+2}ab^{l+3}ab^{l+4}\cdots ab^{m}
\rangle
\end{align*}$$
Take $i, j, k, l$ to be sufficiently large, with each one being at least $n$ orders of magnitude greater than the previous (for some $n>>1$). That the group $G$ is either infinite or torsion-free is not obvious. It relies on something called small cancellation theory. The best reference for this is the book combinatorial group theory by Lyndon and Schupp. The relevant facts are that small cancellation groups are always infinite, and that a small cancellation group contains torsion if and only if a relator is a proper power (and that is not the case here).
To see that the group $G$ has odd-order abelianisation, notice that if we kill the generators $a$ and $b$ we are left with a finite cyclic group. So that is the plan: kill $a$ and $b$. We will assume that the following hold (these assumptions do not affect the infinite-ness nor the torsion-free-ness of the group).


*

*$k-j=jp$.

*$m-l=(l-k)q$.

*$s_k=\sum_{t=1}^{k-j} j+t$ and $s_m=\sum_{t=1}^{m-l} l+t$ are coprime.

*$j$ and $l-k$ are coprime.
Now, the last four relations give four relators in the abelianisation of the form $a^rb^s$, where $r$ is one of $j$, $k-j$, $l-k$, $m-l$. As $k-j=jm$ with $j$ even, we see that $b$ has order dividing $s_k$ in the abelianisation (by combining $a^jb^{s_j}$ with $a^{k-j}b^{s_k}$). Similarly, the relators $a^{l-k}b^{s_l}$ and $a^{m-l}b^{s_m}$ combine to imply that $b$ has order dividing $s_m$. As $s_k$ and $s_m$ are coprime, $b$ is trivial in the abelianisation. Thus, $a$ has order equal to $\gcd(j, k-j, l-k, m-l)$ in the abelianisation. As $j$ and $l-k$ are coprime, $a$ is trivial in the abelianisation. Hence, the abelianisation of the group $G$ is cyclic of odd order, as required.
