The equation $2x=a$ has no solution in an infinite additive cyclic group generated by $a$

Question

Q: If $G$ is an infinite additive cyclic group with the generator element $a$, prove that the equation $2x=a$ has no solution in $G$, $x \in G $.

Answer:

Since $G$ is an infinite cyclic group then $|a|= \infty$ (where $|a|$ denotes the order of $a$)

Since $x \in G $ there for $x=ia$

$$2x=a \Rightarrow 2ia=a$$

$$a(2i-1)=0$$

i'm stuck here so i don't know how to prove that this is unsolvable

Answer 1

You have shown that $\lvert a\rvert$ divides the absolute value of $2i-1$. But the order of $a$ is infinite.

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Alternatively, $G\cong\Bbb Z$ is free of rank one and so has no torsion (but this jumps the gun a bit).

Answer 2

$(2i-1)a=0$ implies the order of $a$ is finite. That's a contradiction.

Answer 3

Using additive notation: if $a$ generates the group $G$ and $2x = a$ for some $x \in G$, then, because $a$ generates $G$, $x = ma$ for some $m \in \Bbb{Z} \setminus \{0\}$, giving us that $2ma = a$. So $G$ cannot be infinite, since we must have $(2m-1)a = 0$.