Algebraic Structure of the Rose with Two Petals I am trying to determine whether the rose with two petals ($S^1 \vee S^1$ or the figure-eight) has a continuous multiplication with identity element.  I know that this is true for the unit circle $S^1$ in the complex plane, where $S^1 = \{ z \in \mathbb{C} \mid |{z}| = 1 \}$.
I also know that $S^1$ is a Lie Group, and I believe that because of the intersection point of the figure eight, this space does not have a continuous multiplication with identity element and is therefore not a topological group.  Would someone mind pointing me in the right direction for how to approach this idea?
 A: The following lemma will be useful:
Lemma. Let $M$ be a topological space with a continuous multiplication $*$ and an identity element $e$. Then $\pi_1 (M)$ is an abelian group. 
Proof. Let $u, v : [0, 1] \to M$ be continuous loops based at the identity element $e$.  We construct a homotopy from $u \mathbin{.} v$ to $v \mathbin{.}  u$ as follows: let $H : [0, 2] \times [0, 2] \to G$ be the map defined by
$$H(s, t) = \begin{cases} 
u(s) & 0 \le s \le 1, s + t \le 1 \\
v(s - 1) & 1 \le s \le 2,  s - t \ge 1 \\
v(s) & 0 \le s \le 1, t - s \ge 1 \\
u(s - 1) & 1 \le s \le 2, s + t \ge 3 \\
u(\tfrac{1}{2}(s - t + 1)) * v(\tfrac{1}{2}(s + t - 1)) & \text{otherwise}
\end{cases} $$
One easily verifies that $H$ is continuous (draw a picture!) and is the required homotopy.
A: In your second paragraph, you are alluding to the idea that $S^1\vee S^1$ may be a Lie group.  To be a Lie group, we require not only a continuous multiplication with identity (which we've already seen $S^1\vee S^1$ can't have), but our space must also be a manifold.  However, $S^1\vee S^1$ isn't.  Indeed, any open neighbourhood of the intersection point in $S^1\vee S^1$ is homeomorphic to a space that looks like an open $+$ sign, which is not homeomorphic to $\mathbb R$.  We can see this because $\mathbb R\backslash \{x\}$ has two connected components for any point $x\in\mathbb{R}$, whereas if $U$ is any neighbourhood of the intersection point $*$ of $S^1\vee S^1$, then $U\backslash \{*\}$ has four connected components.
