# $\langle a_1, \cdots, a_n \mid 2a_1 + \cdots + 2a_n = 0 \rangle \cong \mathbb Z \oplus \cdots \oplus \mathbb Z \oplus \mathbb Z/2$?

Let $$nP$$ be connected sum of $$n$$ Real projective planes. Then, abelianization of $$\pi_1(nP)$$ is the additive group generated by $$a_1, \cdots, a_n$$ with one relation $$2a_1 + \cdots + 2a_n = 0$$, where $$a_1, a_2, \cdots, a_n$$ are generators of $$\pi_1(nP)$$. This makes sense, but why is it isomorphic to: $$\mathbb Z \underbrace{\oplus \cdots \oplus}_{n-1} \mathbb Z \oplus \mathbb Z/2$$?

Can anyone help me to identify such isomorphism? Thank you.

• Smith Normal Form. – Teddy38 Mar 31 at 15:45

## 2 Answers

Let $$v=(1,1,1,\dots,1) \in \mathbb Z^n$$. Then, the vectors $$v,e_2,e_3,\dots,e_n$$ form a basis of $$\mathbb Z^n$$. Your group is isomorphic to $$\mathbb Z^n/ \langle 2v \rangle$$ and so $$\frac{\mathbb Z^n}{\langle 2v \rangle} = \frac{\mathbb Z v \oplus \mathbb Z e_2 \oplus \cdots \oplus Z e_n}{2\mathbb Z v \oplus 0 e_2 \oplus \cdots \oplus 0 e_n} \cong \mathbb Z/2\mathbb Z \oplus \mathbb Z \oplus \cdots \oplus Z$$

$$\pi^{ab}(nP)=\bigoplus_{i=1}^n \mathbb Za_i/ \langle 2\sum_{j=1}^n a_j \rangle$$
Consider the abelian group isomorphism $$\bigoplus_{i=1}^n \mathbb Za_i\rightarrow \bigoplus_{i=1}^n \mathbb Za_i$$ given by $$a_i\mapsto a_i$$ for $$i and $$a_n \mapsto \sum_{j=1}^n a_j$$. Under this isomorphism, if we write $$e_i$$ as the image of $$a_i$$, then
$$\pi^{ab}(nP)=\bigoplus_{i=1}^n \mathbb Za_i/ \langle 2\sum_{j=1}^n a_j \rangle \cong \bigoplus_{i=1}^n \mathbb Ze_i/ \langle 2 e_n \rangle\cong \mathbb Z \underbrace{\oplus \cdots \oplus}_{n-1} \mathbb Z \oplus \mathbb Z/2$$