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Let $F_n$ be the free group with $n$ generators $\{ x_1 , \dots , x_n \}$. From my understanding, we say that a subgroup $K \leq F_n$ is a free factor of $F_n$ if there exists a subgroup $H \leq F_n$ such that $F_n = H * K$. Furthermore, if I understand correctly, if we write that $K = \langle f_1 , \dots , f_s \rangle$ and $H = \langle g_1 , \dots , g_m \rangle$, then we have that $H * K = \langle f_1 , \dots , f_s , g_1 , \dots , g_m \rangle$.

Now my confusion is as follows - with this definition, isn't EVERY subgroup of a free group a free factor? For a silly example, let $n=2$, $F_2 = \langle a, b \rangle$ and let $K = \langle aba , b \rangle$. Then can't we just take $C = \langle a \rangle$, which gives $K * C = \langle aba, b , a \rangle = \langle a, b \rangle = F_2$ ? Isn't it always possible to do this for any $K$, by just letting $C$ to be the subgroup of $F_n$ generated by the $x_i$'s that don't occur in $K$ (so possibly $C = F_2$)?

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Given a group $G$ with subgroups $K$ and $H$, we denote by $G=K * H$ that $G$ is the internal free product of $G$ and $H$. This equivalent to the following two properties being satisfied:

  1. Any element of $G$ can be written as finite product with factors taken from $K$ and $H$.
  2. Any finite product of non-identity factors alternating between $K$ and $H$ is not the identity in $G$.

Alternatively, you can define this as the homomorphism $K*H\to G$ being an isomorphism, where $K*H$ denotes the external free product and the map sends alternating words to the corresponding product in $G$. Surjectivity of this homomorphism is equivalent to (1), while injectivity is equivalent to (2).

When $G=\langle a,b\rangle$ is the free group on two generators, $H=\langle aba,b\rangle$ and $K=\langle a\rangle$, the first property is satisfied, but the second is not: $$ (aba)(a^{-1})(b^{-1})(a^{-1}) = 1 $$ and $aba\in H$, $a^{-1}\in K$, $b^{-1}\in H$ and $a^{-1}\in K$ are all non-identity elements.

In terms of the map on the external free product, the homomorphism $K*H\to G$ would be surjective but not injective in this example, since the alternating word $(aba)(a^{-1})(b^{-1})(a^{-1})$ is not the identity in the external free product $K*H$ but its image in $G$ is. Hence, the kernel is non-trivial.

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  • $\begingroup$ That clarified everything, thank you very much! $\endgroup$ – dacian98 Jan 19 at 17:55
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There's an (understandable) confusion between the elements $aba$, $b$, and $a$ generating $F_2$, versus generating $F_2$ freely (that is, with no relations among them). The notation $H*K$ is a free product, so no relations are allowed between their elements; however, there's an obvious relation among $aba$, $b$, and $a$.

For an even simpler example, consider $F_1 \cong \Bbb Z$, and let $K=\langle 2\rangle$. Then we can similarly ask, can't we just take $C=\langle 5\rangle$, since $\langle 2,5\rangle = \Bbb Z$? No, because $\Bbb Z$ is not the free product of $\langle 2\rangle$ and $\langle 5\rangle$.

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  • $\begingroup$ Does the "free" part of the definition have to do with the universal property of free groups? I'm looking for a way to check whether relatively simple subgroups of $F_2$ are free factors $\endgroup$ – dacian98 Jan 19 at 17:18
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    $\begingroup$ This is kind of tautological, but... the "free" in "free group" does indeed have to do with the universal property of free groups. On the other hand, the "free" in "free product" has to do with the universal property of free products. $\endgroup$ – Lee Mosher Jan 19 at 22:18

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