# Free products of two or copies of a group

Could someone describes the elements of the free product of two (or three...) copies of the finite cyclic group $\mathbb{Z}/N\mathbb{Z}$. Are they words over $\mathbb{Z}/N\mathbb{Z}$ ? Do you allow $0$ (I mean the neutral element of $\mathbb{Z}/N\mathbb{Z}$) in these words? What is the difference from the free product of two and the free product of three copies then?

You may see I'm lost... Thank you...

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Let $a$ be a generator of the first copy of $\mathbb{Z}/N\mathbb{Z}$ and $b$ be a generator of the second copy.

An element in the free product of the two groups is then a word of the form $a^{i_1}b^{k_1}a^{i_2}b^{k_2}\cdots a^{i_m}b^{k_m}$ for some $m$ and with the $i_j$ and $k_j$ between $0$ and $N-1$. If we further only allow $i_1$ and $k_m$ to be $0$, any element has a unique representation of this form.

The important thing to note is that we need to distinguish the generators coming from the two groups, so if we included a third copy, we would have to introduce a third generator as well.

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Ok thank you... – fft Jan 9 '13 at 14:24

They are words over the disjoint union of two (or three) copies of the set of nonzero elements of $\mathbb Z/n\mathbb Z$, with the additional restriction that two neighboring symbols must not come from the same of the two (or three) copies.

In the case of two factors, once we know whether the first symbol is from one factor or the other, the additional restriction makes it unambiguous which one each of the following symbols come from, so in that case the elements of the group are almost just words of nonzero elements of the original group (but plus information about whether it is first-factor-first or second-factor-first). With more than two factors this becomes unworkable and you need to fall back on the "neighboring symbols must not come from the same copy" rule.

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Ok so an element in $Z/NZ*Z/NZ*Z/NZ$ could be $a_1b_1c_1a_2b_2b_3a_4\dots$ with the $a$'s from the first copy ($b,c$ form the second, third ) ? (i.e. $c_2=0=a_3=c_3$... are discarded from the final writing ?). – fft Jan 9 '13 at 14:29
Oh sorry, then the two neighboring b's $b_2b_3$, would reduce to $b_2+b_3$ ? – fft Jan 9 '13 at 14:33
@fft: Yes, at least if you want unique representations of the free product elements. The $b_2b_3$ would reduce to a single element, and if that happens to be the identity, it would disappear and $a_2a_4$ would then reduce to a single element and so forth... – Henning Makholm Jan 9 '13 at 14:43
Ok I think I understood... Thank you – fft Jan 9 '13 at 14:46