Free abelian group $F$ has a subgroup of index $n$?

Suppose that we have a free abelian group $F$. How can it be proved that $F$ has a subgroup of index $n$ which $n≥1$?

Honestly, according to the Theorems, I just know that if we take $X$ as a base for $F$, then $$ F= \bigoplus_{\alpha \in X} \mathbb Z_\alpha \ $$ in which for all $ \alpha \in X$; $\mathbb Z_\alpha \ $ is a copy of $ \mathbb Z $. What that subgroup could be? Thanks.

1
2022-07-25 20:47:13
Source Share
Answers: 0