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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.

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    $\begingroup$ Smith Normal Form. $\endgroup$ – Teddy38 Mar 31 at 15:45
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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 $$ \frac{\mathbb Z^n}{\langle 2v \rangle} = \frac{\mathbb Z v \oplus \mathbb Z e_2 \oplus \cdots \oplus Z e_n}{2\mathbb Z v \oplus 0 e_2 \oplus \cdots \oplus 0 e_n} \cong \mathbb Z/2\mathbb Z \oplus \mathbb Z \oplus \cdots \oplus Z $$

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$\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 $i<n$ 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
$\pi^{ab}(nP)=\bigoplus_{i=1}^n \mathbb Za_i/ \langle 2\sum_{j=1}^n a_j \rangle \cong \bigoplus_{i=1}^n \mathbb Ze_i/ \langle 2 e_n \rangle\cong \mathbb Z \underbrace{\oplus \cdots \oplus}_{n-1} \mathbb Z \oplus \mathbb Z/2$

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