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Show the following: Let $G_1, G_2$ be groups and denote by $G_1*G_2$ their free product. Then center of $G_1*G_2$ contains only the neutral element.

Have you some nice references about free products and groups?

Note that a center of a group contains all elements, which commutes with all elements of the group.

My ideas: Let $g_1 \in G_1, g_2 \in G_2$ and $h$ be an element of the center. It applies $g_1 \cdot h = h \cdot g_1$ and $g_2 \cdot h = h \cdot g_2$. Since $g_1 \cdot h$ and $h \cdot g_1$ are different strings, the only way that $g_1 \cdot h = h \cdot g_1$ could apply is that $g_1 \cdot h$ is not reduced. Analogously we see that $g_2 \cdot h$ is not reduced. It follows that $h$ can't be an element of $G_1$ and $G_2$, because otherwise $g_1\cdot h$ and $g_2 \cdot h$ could not be simultaneously unreduced. So the only way that $g_1\cdot h$ and $g_2 \cdot h$ are simultaneously unreduced, is that $h$ is equal to the neutral element.

Is that a correct proof?

For me seems the proof quite complicated and unprecise, especially the part, where I argue that $g_1 \cdot h$ is not reduced.

Is there a proof which is simpler and clearer?

Bests

bjn

edit 1: Thanks, I replaced "centralizer" by "center"

bjn
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  • The usual name is "the center of the group". Your beginning of proof looks fine, but you could take reduced words in normal form from the beginning and simplify it even further. – Timbuc Nov 02 '14 at 16:25
  • @Timbuc : Can you explain what you mean by "reduced words in normal form"? The introduction I read about free products was quite unprecise and informal. Have you also some good reference for free products and free groups? – bjn Nov 02 '14 at 16:39
  • If you don't tell I can't guess what introduction you're reading (a link will be great here). There are several great texts in this, with classics like Magnus-Karrass-Solitar, or Lyndon-Schupp. Both books are called "Combinatorial Group theory" and in both chapter 4 deals with free products. L-S calls the reduced form "normal" whereas M-K-S only calls it "reduced form". – Timbuc Nov 02 '14 at 17:04
  • It was introduced in the lecture and I tried to read it in Lang's algebra book. I suppose you mean by "reduced normal form", that $h=h_{a_1}\cdot h_{a_2}....\cdot h_{a_m}$, where $a_i \neq a_j$ for any $j,i$ and $h_{a_i}$ is unequal to the neutral element for any $j$. But how this helps? – bjn Nov 02 '14 at 18:50

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