What is $\displaystyle \pi_1(S^1 \vee S^1 \vee S^2)$?

Is it $\displaystyle \mathbb{Z} * \mathbb{Z} * \{1\}$ by Van Kampen's Theorem?

Thanks, I just wanted to check this.

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- May 2nd 2009, 09:50 AMErdos32212Fundamental Group Question
What is $\displaystyle \pi_1(S^1 \vee S^1 \vee S^2)$?

Is it $\displaystyle \mathbb{Z} * \mathbb{Z} * \{1\}$ by Van Kampen's Theorem?

Thanks, I just wanted to check this. - May 2nd 2009, 01:30 PMGaloisTheory1
Yes, that is correct. It is a free abelian group on two generators.

- May 3rd 2009, 04:29 AMaliceinwonderland
Yes.

Since $\displaystyle S^2$ is simply connected, $\displaystyle \pi_1(S^1 \vee S^1 \vee S^2) \cong \mathbb{Z} * \mathbb{Z}$.

The free product of a free group $\displaystyle \mathbb{Z}$ and a free group $\displaystyle \mathbb{Z}$ is a free group having a presentation $\displaystyle \{\alpha, \beta\}$, where $\displaystyle \alpha$ = 1 or -1 in the first $\displaystyle \mathbb{Z}$ and $\displaystyle \beta$ = 1 or -1 in the second $\displaystyle \mathbb{Z}$.

For instance,

$\displaystyle 1_1 1_1 1_1 1_2 1_2$ (followed by a member in the first group) .... ,

reduces to

$\displaystyle 3_1 2_2 ...$ ,

where $\displaystyle x_i$ denotes the element x in the i-th $\displaystyle \mathbb{Z}$, i=1 or 2, and the group operation between elements in the same group is an addition.

As you know, $\displaystyle \mathbb{Z} * \mathbb{Z}$ and $\displaystyle \mathbb{Z} \times \mathbb{Z}$ are two different things. The former is non-abelian because a free group on a nonempty set S is abelian iff S has exactly one element. The latter group is abelian, which is isomorphic to the fundamental group of a torus having a presentation $\displaystyle \{\alpha, \beta\ | \alpha\beta\alpha^{-1}\beta^{-1}=1\}$.