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I was working in an abelian group and came across two elements of order 4 and wanted to know what subgroup they generated. Denote the two elements $x$ and $y$. I also proved $x+y=0$. So the equations we have are: $4x=4y=0$ and $x+y=y+x=0$. Now if I look at $\langle x,y\rangle$, I can only come up with 7 different elements: $\{0,x,2x,3x,y,2y,3y\}$. I cannot think of any other elements, which means $|\langle x,y\rangle| = 7$. As we all know by basic group theory: the order of any element divides the order of the group. This clearly gives a contradiction.

I am sure there does not exist any other element besides those 7 I gave. But this must give a contradiction. I do not think it shows that these $x$ and $y$ cannot exist, since you can also consider the free group of 2 generaters and quotient by the correct equations (which would be $\langle a,b | 4a = 4b = 0, a+b=b+a=0\rangle$).

Is there something I am missing? I really dont see it.

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    $\begingroup$ You are mixing additive and multiplicative notation within your group. That's very confusing, and may explain some of your difficulties. $\endgroup$ Jul 31 '16 at 10:20
  • $\begingroup$ I edited it, only using additive notation from now on! Good point. Also for what my difficulty is: how can $\langle x,y\rangle$ have order 7? $\endgroup$
    – Rico
    Jul 31 '16 at 10:24
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    $\begingroup$ Better now. If $4x=0$ and $x+y=0$, then also $y-3x=(x+y)-4x=0-0=0$. Therefore $y=3x$. Continue and conclude that there are only four elements in the group $x$ and $y$ generate. $\endgroup$ Jul 31 '16 at 10:25
  • $\begingroup$ Wauw that was a really easy answer. You're right, mixing up additive and multiplicitive notation was probably my problem! $\endgroup$
    – Rico
    Jul 31 '16 at 10:27
  • $\begingroup$ To get more feedback you may consider typing it all up as an answer. $\endgroup$ Jul 31 '16 at 10:28
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The subgroup $\langle x,y\rangle$ can only have 7 elements at most which I state in the question. Also since $x$ has order 4 it must be a multiple of 4. A logical conclusion would be that the order is 4. This implies that $x$ and $y$ are related in some way (assume $x\not=y$). As stated in the comments, we have: $$y-3x = (x+y)-4x = 0-0 = 0$$ Therefore we can conclude $y=3x$. This is equivalent to $x=-y$, since $\overline{3} = \overline{-1} \in \mathbb{Z}/4\mathbb{Z}$.

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