Does abelianization change the rank of $\pi_1(X)$?

I am considering the relationship in between the first homology team $H_1(X)$ and also the basic team $\pi_1(X)$. Offered a course - linked room $X$, I have some grip of what $\pi_1(X)$ need to be with ease (regarding I realize, it can be informally be taken the variety of 'openings' in the room). Hence, it would certainly be a really simple to reason $H_1(X)$ if I recognized that taking the abelianization $\pi_1(X)_{ab}$ does not transform the ranking of the basic team.

As an instance, $S^1 \vee S^1$ has 2 'openings', therefore with ease one could assume that $\pi_1(X) \cong \mathbb{Z} \oplus \mathbb{Z}$. It ends up additionally that $H_1(X) \cong \mathbb{Z} \oplus \mathbb{Z}$, so they have the very same ranking.

My inquiry is, does this constantly hold? Exists some instance where the ranking of $H_1(X)$ and also $\pi_1(X)$ are various?

2022-06-07 14:40:55
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Answers: 1

If $G$ is a team, after that $G$ and also $G_{ab}$ require not be of the very same ranking. As an example, $A_5$ has ranking 2, while its abelianization is the unimportant team, with ranking 0. Every team is the basic team of some path - linked room, so this reveals there is a room $X$ such that $H_1(X)$ and also $\pi_1(X)$ have various rankings.

2022-06-07 15:09:09