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Reading the book "A Course in the Theory of Groups" by D. J. S. Robinson, I was looking at the proof of 6.4.3 (iii), which states (suppose we are in the case of two groups): if $G_1$ and $G_2$ are groups, and $H$ is (isomorphic to) a subgroup of every $G_\lambda$, let $G:=G_1*_HG_2$ be their amalgamated free product; then an element of $G$ has finite order if and only if it is conjugate to an element of some $G_\lambda$.

We know that $G_1*_HG_2\cong\frac{G_1*G_2}{N}$ where $N$ is the normal subgroup of $G_1*G_2$ generated by $\{ h\theta (h)^{-1}\, |\, h\in H\}$. Therefore any element of $G_1*_HG_2$ is in fact of the form $gN$ for $g\in G_1*G_2$.

My question is: in the statement, are we taking the conjugate in $G:=G_1*_HG_2$? In other words, does the statement mean that for every $gN$ of finite order in $\frac{G_1*G_2}{N}$ there exists $w\in G_1*G_2$ and $a\in G_\lambda$ (for some $\lambda\in \{ 1,2\})$ such that $w^{-1}gwN =aN$?

If this is the case, I don't get the proof of the book, since to me it seems that the book proves a stronger condition: I think that for every $gN$ of finite order in $\frac{G_1*G_2}{N}$ there exists $a\in G_\lambda$ (for some $\lambda\in \{ 1,2\})$ such that $gN =aN$. This is a stronger condition, and I don't see why the books state the theorem with conjugation (also "An Introduction to the Theory of Groups" by J. J. Rotman does, in Theorem 11.66).

Indeed, the proof shows that the only normal form in the class $gN$, which is of the form $h\bar{g}_1\cdots\bar{g}_n$, has necessarily $n=1$ (otherwise we get a contradiction, as by conjugating we get a finite order $g'N$ whose associated normal form has length reduced by 1), i.e. $gN=h\bar{g}_1N$, where $\bar{g}_1\in G_{\lambda_0}$ for some $\lambda_0\in\{1,2\}$. This clearly implies that, by identifying $h$ with its image in $G_{\lambda_0}$, $h\bar{g}_1$ is a finite order representative of $gN$ and belongs to $G_{\lambda_0}$.

Therefore: am I missing some subtlety in the proof, or is the statement meant in another way? For example, maybe the statement tells that the representative $g\in G_1*G_2$ is conjugate in $G_1*G_2$ to some $a\in G_{\lambda_0} $ for some $\lambda_0\in\{ 1,2\}$? But it would not be clear to me how this would follow from the same proof.

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  • $\begingroup$ The answer is in your second to last paragraph. If n is not 1 you conjugate to decrease length. So for an example of minimal length it belongs to one of the factors and you can get to minimal length by conjugating. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 21, 2023 at 1:39
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    $\begingroup$ The more elegant way to prove this is to use Bass-Serre theory. The amalgamated product acts on a tree without inversions with two orbits of vertices one stabilized by G1 and one by G2. Any finite group acting on a tree fixes a vertex and so your element of finite orbit is conjugate to an element of G1 or G2 $\endgroup$
    – Benjamin Steinberg
    Commented Nov 21, 2023 at 1:43
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    $\begingroup$ Just look at the free product of two groups of order 2 generated by a,b. Then aba has order 2 but is not in either factor. But it is conjugate to b. More generally abababababa has order 2 and is conjugate to b. $\endgroup$
    – Benjamin Steinberg
    Commented Nov 21, 2023 at 2:36
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    $\begingroup$ To define the amalgamated product $G_1\ast_H G_2$, it is not enough to assume that $H$ is isomorphic to subgroups of $G_1$ and $G_2$. You need to fix such isomorphisms. $\endgroup$
    – YCor
    Commented Nov 21, 2023 at 8:03
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    $\begingroup$ It is clear that all elements of a conjugacy class have the same order, so once one element of a conjugacy class has finite order, all elements of that class do. I think the proof of the result you want is given very clearly in Serre's book on Trees ( presumably the same as the one @BenjaminSteinberg alludes to). $\endgroup$
    – Geoff Robinson
    Commented Nov 21, 2023 at 11:52

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