I'm new to the idea of a free product..
Basically I was wondering if G is an arbitrary group and 1 is the trivial group then is $1\star G \cong G$. If not.. what whould it look like?
[Math] Free product of the trivial group with another group
abstract-algebraalgebraic-topologyfree-groupsgroup-theory
Related Solutions
Yes, the free product of the free group of rank $n$ and the free group of rank $m$ is isomorphic to the free group of rank $n+m$. Note also that if $G$ is free of rank $2$, and two elements $a$ and $b$ generate $G$, then $a$ and $b$ must freely generate $G$. One way to see this is to invoke the fact that a free group of finite rank is Hopfian. Another is to assume there is a nontrivial reduced word in $a$ and $b$ equal to the identity, take a free generating set $x$ and $y$, express $a$ and $b$ in terms of $x$ and $y$, and replace them in the nontrivial word expressing the identity; this will yield a nontrivial word in $x$ and $y$ equal to the identity (some work needs to be done here, of course), contradicting the choice of $x$ and $y$ as a free generating set. And there are other ways, of course.
An easy way to get this is as a corollary to the fact that the free group functor is the left adjoint of the underlying set functor. That is, for every group $G$ and every set $X$, $$\mathcal{G}roup(\mathbf{F}(X),G) \longleftrightarrow \mathcal{S}et(X,\mathbf{U}(G)),$$ where $\mathbf{F}(X)$ is the free group on the set $X$ and $\mathbf{U}(G)$ is the underlying set of the group $G$.
Because the free group functor is a left adjoint, it sends coproducts to coproducts. That is, the coproduct of two free groups $F(X)$ and $F(Y)$ in the category of groups is the free group on the coproduct of $X$ and $Y$ in the category of sets. The coproduct in the category of groups is the free product, and the coproduct in sets is the disjoint union. Therefore, there is a natural isomorphism $$F(X\amalg Y) \cong F(X)*F(Y).$$
You can also prove it directly from the universal property: the universal property of the free group on $X\amalg Y$ (the disjoint union) is that for every set-theoretic map $f\colon X\amalg Y\to G$ to a group $G$, there is a unique group homomorphism from $F(X\amalg Y)\to G$ that extends $f$. On the other hand, the universal property of the free product $F(X)*F(Y)$ is that for every pair of group homomorphisms $\varphi\colon F(X)\to G$ and $\psi\colon F(Y)\to G$, there is a unique group homomorphism $\Psi\colon F(X)*F(Y)\to G$ such that $\Psi\circ i_{F(X)}=\varphi$ and $\Psi\circ i_{F(Y)}=\psi$, where $i_{F(X)}\colon F(X)\to F(X)*F(Y)$ and $i_{F(Y)}\colon F(Y)\to F(X)*F(Y)$ are the canonical inclusions.
A set-theoretic map $f\colon X\amalg Y\to G$ is equivalent to a pair of maps $g\colon X\to G$ and $h\colon Y\to G$; the map $g\colon X\to G$ induceds a map $\varphi\colon F(X)\to G$, while the map $h\colon Y\to G$ induces a map $\psi\colon F(Y)\to G$, which in turn induces a map $F(X)*F(Y)\to G$; it is now straightforward to verify that this map extends $f$, and that it is unique, so that $F(X)*F(Y)$ has the universal property of $F(X\amalg Y)$, and therefore they are isomorphic.
Or: the inclusions $X\to F(X)\to F(X)*F(Y)$ and $Y\to F(Y)\to F(X)*F(Y)$ induce an inclusion $X\amalg Y\to F(X)*F(Y)$, which in turn induces a morphism $F(X\amalg Y)\to F(X)*F(Y)$. Conversely, the map $X\to F(X\amalg Y)$ induces a map $F(X)\to F(X\amalg Y)$, and $Y\to F(X\amalg Y)$ induces a map $F(Y)\to F(X\amalg Y)$, and these two maps together induce a map $F(X)*F(Y)\to F(X\amalg Y)$. It is now easy to verify that the induces maps $F(X\amalg Y)\to F(X)*F(Y)$ and $F(X)*F(Y)\to F(X\amalg Y)$ are inverses of each, in the usual abstract nonsense argument about their compositions having the same universal property as the corresponding identity.
P.S. The equality $F(a,b)=\langle a\rangle*\langle b\rangle$ presumably just means that there is a unique isomorphism between the two objects that maps $\{a\}$ and $\{b\}$ to themselves as the identity; this is a consequence of their respective universal properties/adjointness of free group construction.
Let $F_1,F_2$ be non-cyclic finitely generated groups. Then $F_1\times F_2$ is not amalgamated product or HNN extension over a cyclic subgroup.
Let $F_1,F_2$ be non-cyclic finitely generated groups. Suppose that $G=F_1\times F_2$ acts unboundedly, inversion-free, on a tree $T$ with cyclic edge stabilizers.
Suppose that each element of $F_1$ acts with a fixed vertex. Then $F_1$ has a fixed vertex (this is a general fact on tree actions of f.g. groups). Let $T_1$ be the set of vertices fixed by $F_1$. Then $T_1$ is a subtree, and is $F_2$-invariant. By minimality, we deduce that $T=T_1$, and hence $F_1$ acts trivially. Since $F_1$ is not cyclic, it follows that edge stabilizers are not cyclic, contradiction.
So some element $g_1$ of $F_1$ is loxodromic, with axis $A_1$. Similarly, some element $g_2$ of $F_2$ is loxodromic, with axis $A_2$. Since $F_2$ commutes with $g_1$, $F_2$ preserves $A_1$, and hence $A_2=A_1$, and since $F_1$ commutes with $g_2$, $F_1$ preserves $A_2=A_1$. So the action preserves an axis, hence by minimality $T$ is reduced to an axis.
Hence $G$ modulo a cyclic normal subgroup is trivial, of order $2$, infinite cyclic, or dihedral. This is excluded by the assumptions.
Best Answer
Since you tagged it algebraic-topology, perhaps you are learning about free products in a topology course? In this case, if $X$ is a space with fundamental group $G$, then $1*G$ is the fundamental group that you get of the space $X$ with a point glued to a point of $X$, which is just isomorphic to $X$ again, so $1*G\cong G$.