Take the infinite abelian subgroup of $GL_2(\mathbb{R})$, $N=\left\{\textstyle\begin{pmatrix}\lambda &0\\0&\lambda\end{pmatrix}:\lambda \in \mathbb{R}^+\right\}$. Do all subgroups of $N$ also have infinite order?
If so why? And how can I show that?
2026-03-31 20:55:15.1774990515
Order of the subgroups of an infinite abelian group
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That is not true. For example {I} is a definite subgroup of N. If it is about proper subgroups It's true. You can prove in diffrent ways. One way: Proof by contradiction Assume that there is a proper subgroup which has definite order. Beacuse it is proper subgroup, it has a member which is not equal to identity. While productions of this member is shown as lambda to n, you have to show for all integers for n, these powers of lambda are not equal So infinite members are creating by producting .