The following statement is given in the third comment of kernel of the mod $2$ Bockstein on the first cohomology group:
Statement: Let $X$ be a path-connected finite $CW$-complex. Suppose $$ H_1(X;\mathbb{Z})=\mathbb{Z}_2^{\oplus r}\oplus A $$ where $r\geq 0$ and $A$ is a finite abelian group of odd order. Then for any nonzero element $x\in H^1(M;\mathbb{Z}_2)$, $x^2\neq 0$.
My attempt to prove the statement: I notice that for any nonzero element $x\in H^1(M;\mathbb{Z}_2)$, $x^2=Sq^1 x=\beta x$ where $\beta$ is the Bockstein homomorphism associated with the coefficient sequence $$ 0\to \mathbb{Z}_2 \to\mathbb{Z}_4 \to \mathbb{Z}_2\to 0.$$ Hence we only need to prove $$ \text{Ker} \beta=0. $$ By the universal coefficient theorem, $$ H^1(M;\mathbb{Z}_4)=Hom(H_1(M;\mathbb{Z});\mathbb{Z}_4) =Hom(\mathbb{Z}_2^{\oplus r};\mathbb{Z}_4) =\mathbb{Z}_2^{\oplus r},$$
$$ H^1(M;\mathbb{Z}_2)=Hom(H_1(M;\mathbb{Z});\mathbb{Z}_2) =Hom(\mathbb{Z}_2^{\oplus r};\mathbb{Z}_2) =\mathbb{Z}_2^{\oplus r}.$$ By the construction of Bockstein homomorphism, we have an exact sequence $$ H^1(M;\mathbb{Z}_4)\overset{f}{\longrightarrow} H^1(M;\mathbb{Z}_2)\overset{\beta}{\longrightarrow }H^2(M;\mathbb{Z}_2).$$
Hence when $r=0$, I obtain $$ \text{Ker}\beta=\text{Im} f=0. $$
Whether is my above argument right?
Question: When $r>0$, could we still obtain the statement, which is equivalent to prove that the image of $f$ is zero?
The map $f$ isn't just any map: it's the map induced by the quotient map $q:\mathbb{Z}_4\to\mathbb{Z}_2$ on coefficients. By the naturality of the universal coefficients theorem (with respect to the coefficient group), the map $f:H^1(M;\mathbb{Z}_4)\to H^1(M;\mathbb{Z}_2)$ can be identified with the map $\operatorname{Hom}(H_1(M;\mathbb{Z}),\mathbb{Z}_4)\to\operatorname{Hom}(H_1(M;\mathbb{Z}),\mathbb{Z}_2)$ given by taking a homomorphism $H_1(M;\mathbb{Z})\to\mathbb{Z}_4$ and composing it with $q$ to get a homomorphism $H_1(M;\mathbb{Z})\to\mathbb{Z}_2$. But because $H_1(M;\mathbb{Z})\cong \mathbb{Z}_2^r\oplus A$, the image of every homomorphism $H_1(M;\mathbb{Z})\to\mathbb{Z}_4$ is contained in the kernel of $q$ (since every element not in the kernel of $q$ has order $4$, but $H_1(M;\mathbb{Z})$ has no elements of order divisible by $4$). So the map $\operatorname{Hom}(H_1(M;\mathbb{Z}),\mathbb{Z}_4)\to\operatorname{Hom}(H_1(M;\mathbb{Z}),\mathbb{Z}_2)$ is identically $0$.