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As Vahid says, it is true for any topological group. Here is a proof. I'm sure there are nicer, more conceptual ones out there, but here goes.

Let $G$ be your topological group. Take two loops $\sigma$ and $\gamma$ in $G$, based at the identity of $G$, which we will denote by $e$. Let $\sigma \cdot \gamma$ be the concatenation of the two loops. This is given by $$(\sigma \cdot \gamma) (t) = \begin{matrix} begin{cases} \sigma(2t) & \quad \text{ if } 0 \le t \le 1/2 \\ \gamma(2t-1) &\quad \text{ if } 1/2 \le t \le 1 \end{matrix} end{cases}$$ (Sorry, couldn't manage to format that any better. Feel free to edit if you know how to put a nice brace bracket to the left of that definition.)

The idea is this. We will show that $\sigma \cdot \gamma$ is homotopic to to the loop given by the pointwise product of $\sigma$ and $\gamma$. Let's call that loop $\rho$, so $$\rho(t) = \sigma(t)\gamma(t).$$

Now define an auxiliary function $P : [0,1] \times [0,1] \to G$ by $$P(s,t) = \begin{matrix} begin{cases} \sigma\left( \frac{2t}{1+s} \right) & \quad \text{ if } 0 \le t \le \frac{1+s}{2} \\ e &\quad \text{ if } \frac{1+s}{2} \le t \le 1 \end{matrix}$$end{cases}$$At s=0, this function does the whole loop \sigma as t goes from 0 to 1/2, then sits at e. In other words, at s=0 this is the first half of the loop \sigma \cdot \gamma. As s gets larger, P does the whole loop \sigma as t goes from 0 to \frac{1+s}{2}. At s=1, P does the loop \sigma at normal speed. Then similarly define a function Q : [0,1] \times [0,1] \to G by$$ Q(s,t) = \begin{matrix} begin{cases} e & \quad \text{ if } 0 \le t \le \frac{1-s}{2} \\ \gamma \left( \frac{2t-1+s}{1+s} \right) &\quad \text{ if } \frac{1-s}{2} \le t \le 1 \end{matrix}$$end{cases}$$

At $s=0$ this is just the second half of the loop $\sigma\cdot\gamma$, while at $s=1$ it is exactly the loop $\gamma$.

So finally, define $$H(s,t) = P(s,t) \cdot Q(s,t).$$ At $s=0$ this is $\sigma \cdot \gamma$, while at $s=1$ it is the pointwise product loop $\rho$. $H$ is clearly continuous, and $H(s,0) = e = H(s,1)$ for all $s$, so this is a homotopy of loops between $\sigma \cdot \gamma$ and $\rho$.

Now we can redo that process and show that $\rho$ is homotopic to the other concatenation $\gamma \cdot \sigma$. So this shows that $\pi_1(G)$ is abelian.

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As Vahid says, it is true for any topological group. Here is a proof. I'm sure there are nicer, more conceptual ones out there, but here goes.

Let $G$ be your topological group. Take two loops $\sigma$ and $\gamma$ in $G$, based at the identity of $G$, which we will denote by $e$. Let $\sigma \cdot \gamma$ be the concatenation of the two loops. This is given by $$(\sigma \cdot \gamma) (t) = \begin{matrix} \sigma(2t) & \quad \text{ if } 0 \le t \le 1/2 \\ \gamma(2t-1) &\quad \text{ if } 1/2 \le t \le 1 \end{matrix}$$ (Sorry, couldn't manage to format that any better. Feel free to edit if you know how to put a nice brace bracket to the left of that definition.)

The idea is this. We will show that $\sigma \cdot \gamma$ is homotopic to to the loop given by the pointwise product of $\sigma$ and $\gamma$. Let's call that loop $\rho$, so $$\rho(t) = \sigma(t)\gamma(t).$$

Now define an auxiliary function $P : [0,1] \times [0,1] \to G$ by $$P(s,t) = \begin{matrix} \sigma\left( \frac{2t}{1+s} \right) & \quad \text{ if } 0 \le t \le \frac{1+s}{2} \\ e &\quad \text{ if } \frac{1+s}{2} \le t \le 1 \end{matrix}$$

At $s=0$, this function does the whole loop $\sigma$ as $t$ goes from $0$ to $1/2$, then sits at $e$. In other words, at $s=0$ this is the first half of the loop $\sigma \cdot \gamma$. As $s$ gets larger, $P$ does the whole loop $\sigma$ as $t$ goes from $0$ to $\frac{1+s}{2}$. At $s=1$, $P$ does the loop $\sigma$ at normal speed.

Then similarly define a function $Q : [0,1] \times [0,1] \to G$ by $$Q(s,t) = \begin{matrix} e & \quad \text{ if } 0 \le t \le \frac{1-s}{2} \\ \gamma \left( \frac{2t-1+s}{1+s} \right) &\quad \text{ if } \frac{1-s}{2} \le t \le 1 \end{matrix}$$

At $s=0$ this is just the second half of the loop $\sigma\cdot\gamma$, while at $s=1$ it is exactly the loop $\gamma$.

So finally, define $$H(s,t) = P(s,t) \cdot Q(s,t).$$ At $s=0$ this is $\sigma \cdot \gamma$, while at $s=1$ it is the pointwise product loop $\rho$. $H$ is clearly continuous, and $H(s,0) = e = H(s,1)$ for all $s$, so this is a homotopy of loops between $\sigma \cdot \gamma$ and $\rho$.

Now we can redo that process and show that $\rho$ is homotopic to the other concatenation $\gamma \cdot \sigma$. So this shows that $\pi_1(G)$ is abelian.