I am encountering free products for the first time in Algebraic Topology during the discussion of van Kampen's theorem, and I can't seem to tell the difference between a free product of groups, and a free group. The definition I know (of a free product) is:

Let $G,H$ be groups. Define $G *H$ to be the set of all formal words $g_1h_1 \cdots g_nh_k$ where $g_i \in G$ and $h_j \in H$. Then $G*H$ is a group under the operation of juxtaposition, and the identity is the empty word.

This definition can be extended to the free product of an arbitrary collection $G_\alpha$ of groups, but I don't see how this definition is different from the free group on a set $S$. My guesses at the differences:

  1. Free groups can be formed with any set $S$, whereas free products are defined for collections of groups.
  2. Free products respects relations between the groups. For example, in $G*H$ the word $g_1g_2h_1$ is just $g_3h_1$ for some $g_3=g_1g_2 \in G$. Whereas there is no underlying relation for words in a free group (except for the formal cancellation of words).

And one last questions: are all free groups also free products? Or is the inclusion in the other direction?

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    $\begingroup$ If you take $\bigstar_{s\in S} \mathbb Z$, you get the free group on $S$. The free product is, in a sense, a generalization. It might be useful to know that the "free product" is actually a categorical "coproduct". $\endgroup$ – Justin Young Feb 18 '16 at 1:39
  • $\begingroup$ Thank you. That makes a lot of sense. I have almost no background in category theory so I'm not quite sure what that means but thanks all the same. $\endgroup$ – Kevin Sheng Feb 18 '16 at 1:41
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    $\begingroup$ a free group is a free product of any amount of factors $\Bbb Z$, but in general a free product takes any class of groups to construct a new group, and furthermore, with respect to the Seifert - van Kampen's construction, there is another product of groups: the amalgamated free product . $\endgroup$ – janmarqz Feb 18 '16 at 1:50
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    $\begingroup$ There is a problem of types, so that your question does not make a lot of sense! A free group is a type of group, while the free product of groups is an operation which given a bunch of groups, gives you a new group. The "difference" you ask for in the title is, then, that the two things are completely different beasts! $\endgroup$ – Mariano Suárez-Álvarez Feb 18 '16 at 1:57

In a free group on a generating set, say $\{a,b\}$ every element can be expressed uniquely as $g_1^{r_1}g_2^{r_2}\cdots g_n^{r_n}$ where $g_i\in\{a,b\}$ and $g_{i+1}\ne g_i$ and each $r_i\in\mathbb Z$. For example, $a^2b^{-3}a^4b^2$. This is not necessarily true for a free product. Let $G\cong\mathbb Z/2\mathbb Z$ and $H\cong\mathbb Z/3\mathbb Z$. Let $G=\langle a\rangle$ and $H=\langle b\rangle$. Here, $a^4b^{-5}=b$.

In general, a free group on $S$ is a free product of the infinite cyclic groups generated by each member of $S$. A free product of groups is not necessarily a free group.

  • $\begingroup$ So finite groups can never be free groups? $\endgroup$ – Kevin Sheng Feb 18 '16 at 2:10
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Tim Raczkowski Feb 18 '16 at 2:18
  • $\begingroup$ @KevinSheng The trivial group is free, but other than that, no. $\endgroup$ – Najib Idrissi Feb 18 '16 at 8:27
  • $\begingroup$ Ah yes, the free group on an empty set of generators. Good point. $\endgroup$ – Tim Raczkowski Feb 18 '16 at 15:23
  • $\begingroup$ Is there a finite group, which is a free product of two non trivial groups $\endgroup$ – MANI Sep 10 at 11:43

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