Free product of the trivial group with another group 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?
 A: 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 $1*G\cong G$.
A: Yes. 
For any group $H$, we have $$\hom(1*G, H) \simeq \hom(1, H) \times \hom(G, H) \simeq \hom(G, H).$$ By the Yoneda lemma, it follows that $1* G \simeq G$.
A: Yes, $1*G\cong G$. A proof would depend on which definition or construction of free products you know or like.  Using the definition as the coproduct in the category of groups, the proof consists of observing that, for every group $H$, there's a bijection between the set of homomorphisms $G\to H$ and the set of pairs of homomorphisms, the first $1\to H$ and the second $G\to H$, because there's exactly one homomorphism $1\to H$.  (More generally, a coproduct of an initial object and an arbitrary object $X$ is isomorphic to $X$.)
A: It's true that $1 * G \cong G$. The trivial group contains only an identity element, which is identified with the identity element of $G$ to form the free product.
A: You are right, $1\star G\cong G$.
For any group $X$ and homomorphisms $1\to X$ and  $G\to X$ there is a unique homomorphism $G\to X$ that makes the required diagrams commute - this is simly so because a homommorphism from $1$ to a group is in no way a restriction or condition.
