Need help understanding what is a free factor of a free group.

Question

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 $C = F_2$)?

Answer 1

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.

Answer 2

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 $\langle 5\rangle$.