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Taken from Hatcher, I think the free product $G =\mathbb{Z}/2\mathbb{Z}\ast\mathbb{Z}/2\mathbb{Z}$ should have infinite elements taking the form of words containing alternate elements, i.e. $a, b, ab, ba aba, bab, abab, baba, ababa \dots$, is that correct?

The next part of the question is about a normal subgroup generated by the word $g_1\ast g_2\ast g_1^{-1} \ast g_2^{-1}$, where $g_1$ is the generator of the first copy of $\mathbb{Z}_2$ and $g_2$ is the generator of the second copy, which to me would just look like $abab$ since $a^2=e$ in the first copy and $b^2=e$ in the second.

How many elements does the group $G/N$ (where $N$ is the normal subgroup described above) have?

I think $G/N$ should look something like this $\{e=abab=baba, a, b, ab, ba, aba, bab\}$ and therefore $G/N$ has 7 elements.

Am I working along the right lines or have I misunderstood something badly here?

Edit: Thank you for the answers, I understand where I went wrong.

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    $\begingroup$ Since $abab\in N$, you have $ab=ba$ in $G/N$. Thus $aba=a^2b=b$. The same goes for $bab=a$. $\endgroup$ – Joe Johnson 126 Apr 18 '15 at 15:54
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  1. Yes, it is.

  2. Notice that

    • $baN=ab(b^{-1}a^{-1}ba)N=abN$

    • $abaN=a(baN)=a(abN)=(a^2)bN=bN$

    • $babN=aN$ for the same reason.

    Therefore, no: $G/N$ is indeed the Klein 4-group.

Added

Notice that if a normal subgroup contains $abab$, then it must contain $a^{-1}(abab)a=baba$...

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$N$ seems to be the commutator subgroup of $G$, and so the quotient should just be the Klein-four group.

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Yes. The free product of more than two *non-trivial_ groups is always an infinite group.

About $\;N=\langle\,g_1g_2g_1^{-1}g_2^{-1}\,\rangle^G=\langle\,(g_1g_2)^2\,\rangle^G\;$ , but why would this equal $\;1\;$ ? It doesn't, as then $\;G\;$ would be finite: you do not have commutativity between elements of the two free factors.

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