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I know that the only groups of order 4 are $\mathbb{Z}_2\times \mathbb{Z}_2$ and $\mathbb{Z}_4$ up to isomorphisms. And I also know that the group presentation of $\mathbb{Z}_4$ is $\left ( a:a^4=1 \right )$.

Naturally, I thought the group presentation of $\mathbb{Z}_2\times \mathbb{Z}_2$ is $\left ( a,b:a^2=1,b^2=1 \right )$. But I'm not sure this presentation really indicates $\mathbb{Z}_2\times \mathbb{Z}_2$.

For example, if $A=\begin{pmatrix} \sqrt{2} &1 \\ -1 &-\sqrt{2} \end{pmatrix}, B=\begin{pmatrix} \sqrt{3} &1 \\ -2 &-\sqrt{3} \end{pmatrix}\in \left ( M(2, \mathbb{R}), \cdot \right )$, then a group $\left ( A,B \right )$ has a presentation $\left ( A,B:A^2=E,B^2=E \right )$ but $AB\neq BA$ and thus this group is not isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_2$.

Where did I misunderstand?

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    $\begingroup$ What is the question? $\endgroup$ – user87543 Dec 24 '13 at 5:23
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    $\begingroup$ The relations $a^2=1$ and $b^2=1$ are not enough to specify $C_2\times C_2$ (they actually specify the free product $C_2*C_2$); one needs the additional relation $ab=ba$. $\endgroup$ – anon Dec 24 '13 at 5:24
  • $\begingroup$ Alternatively, the presentation $(a,b:g^2=1\,\,\forall g)$ works. $\endgroup$ – Ian Coley Dec 24 '13 at 5:28
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You are correct that $\langle a, b | a^2 = b^2 = 1\rangle$ is not enough to specify the given group: In fact, this is the large non-abelian group $\mathbb{Z}_2 \ast \mathbb{Z}_2$, where $\ast$ is the free product. Its elements can be thought of as sequences of various forms: $$abab\dots abab, baba\dots baba, abab\dots aba, baba\dots bab$$ So it's necessary to add a bit more; the easiest way would be to modify the definition to add a commutativity relation:

$$\langle a, b | a^2 = b^2 = 1, ab = ba\rangle$$

Then one can rearrange the given sequences, finding that all elements are of the form $a^k b^j$; then there are exactly four group elements, corresponding to the possible parities of $k$ and $j$.

There are, of course, other ways to represent this group.

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Using your original presentation, imagine writing words: they're all of the form $\ldots abab\ldots$. If you add the commutation relation $ab = ba$ to the presentation, these words collapse to be of the form $a^k b^j$ for $k, j = 0$ or $1$. In particular, there are at most 4 elements to any group that satisfies the presentation $(a, b : a^2 = b^2 = 1, ab = ba)$, and since $\mathbb{Z}_2 \times \mathbb{Z}_2$ satisfies it, you've found its presentation.

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This post can be a complement of @T bonger's post. We know that every finite group has a finite presentation and you know that. But having a finite presentation doesn't give us that the group itself is of finite order necessarily ( See Boger's free product of two cyclic groups). So for not to face these cases, we use an idea called The deficiency of group presentation. In your group, as it is in line $3$, the deficiency is zero so according to Mcdonald's theorem, it is infinite and this is not what you wanted so we should make it finite, so we need additional relation to be added to other relations inside the presentation (Line $3$). What would this new presentation be? As you are working on an abelian group Klein $4$- group so the trivial relation is $ab=ba$.

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