# Van Kampen theorem?

So I'm trying to use Van Kampen theorem to prove that a space is null-homotopic.

The thing is I got it down to this $\langle a\mid a=1\rangle$, however I'm confused what does this mean.

For calculating the the torus you get it down to this $\langle a,b\mid a^{-1}b^{-1}ab=1\rangle \cong \mathbb{Z} \times \mathbb{Z}$.

But, was thinking does the brackets mean $\langle 1\rangle \cong \mathbb{Z}$ . I'm confused as normally the $\langle$ , $\rangle$ brackets means the generating set.

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$\langle a\;|\;a=1 \rangle$ is the group with one generator $a$, which satisfies $a = 1$, hence the trivial group. BTW, what does it mean for a space to be null-homotopic? Did you mean simply connected? – Alexander Thumm Mar 14 '12 at 10:09
In the brackets we have generators G and relations R: <G|R>. If there is no relation and one generator, then this group is isomorphic to $\mathbf{Z}$. – HbCwiRoJDp Mar 14 '12 at 10:10

More in general with Van Kampen theorem one have to deal with notation of the following kind

$$\langle g_1,\dots,g_s \mid m_1(g_1,\dots,g_s) = n_1(g_1,\dots,g_s);\dots; m_k(g_1,\dots,g_s) = n_k(g_1,\dots,g_s) \rangle$$ where $m_i(g_1,\dots,g_s)$ and $n_j(g_1,\dots,g_s)$ are strings containing as symbols $g_i$ or $g_i^{-1}$.

This notation is called a finite presentation of a group: it simply denote the biggest group having $s$ generators, named $g_1,\dots,g_s$ for which the relations

$$m_i(g_1,\dots,g_s) = n_i(g_1,\dots,g_s)$$ hold. More formally the notation above denote the quotient group of the free group with $s$-generators by the smaller normal subgroup containing the elements of the form

$$m_i(g_1,\dots, g_s) \left(n_i(g_1,\dots,g_s)\right)^{-1}\ \text{.}$$

In this group the equalities $m_i=n_i$ hold and every other group for which $m_i=n_i$ is a quotient of this group, thus the attribute bigger.

Some additional notes: in your examples $\langle a \mid a = 1 \rangle$ means the group with one generator with is equal to the identity, the group with this property is the trivial group $\{1\}$ that contain just the identity (so this proves that you space is indeed simply connected. In the case of the torus $\langle a,b \mid aba^{-1}b^{-1} = 1\rangle$ is also the group $\langle a,b \mid ab=ba\rangle$, this group is the biggest group having two generators which commute, i.e. it is the free abelian group with two generators.

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The notation $\langle a|a=1\rangle$ would usually mean the group generated by $a$ subject to the relation $a=1$, so in fact it is the trivial group and your space is null-homotopic as you require. What the notation is really saying is that it's the quotient of the free group on $a$ by itself, which is clearly trivial.

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Here I'm reading "null-homotopic" as "simply connected", i.e. every loop is null-homotopic. Thanks to Alexander Thumm for pointing this out in his comment. – Matthew Pressland Mar 14 '12 at 10:13