Suppose $G = G_1 * G_2$. let $c \in G$ and let $A = cG_1c^{-1}$. Show that $A\cap G_2 = \{1\}$.
I think one must prove from contradiction, so suppose this intersection has another point $x \in G_2$ in the intersection. then:
$x = (g_1,g_2)y(g_1,g_2)^{-1}$ for some $g_1,y \in G_1, g_2 \in G_2$
Here i get stuck though because i don't think one can reduce the word on the right hand side to a word of length 1, unless $y = \{1\}$. but then $x$ must be $\{1\}$?
Kees
Hint. The inclusion homomorphisms $G_1\to G_1\times G_2$ and $G_2\to G_1\times G_2$ extend to a homomorphism $\eta :G_1\ast G_2\to G_1\times G_2$ (i.e., $\eta$ restricts to the identity map on each of $G_1$ and $G_2$). Consider the image of an element in $A\cap G_2$ under $\eta$ and use what you know about the direct product.