How Does a Connected Subspace Generate Another in Topological Groups?

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Here is this week's POTW:

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Show that a connected subspace of a topological group $\pi$ which contains the identity must algebraically generate another connected subspace of $\pi$.
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No one answered this week's problem. You can read my solution below.

Spoiler

Let $C$ be a connected subspace of $\pi$ which contains the identity. In particular, the subgroup $\langle C\rangle$ generated by $C$ contains the identity. Let $x$ be a non-identity element of $\langle C\rangle$. There exists an index $k$, elements $c_1,\ldots, c_k\in C$, and positive integers $n_1,\ldots, n_k$ such that $x = c_1^{\epsilon_1 n_1}\cdots c_k^{\epsilon_k n_k}$, where the $\epsilon_i$ are $\pm 1$. Since multiplication and inversion are continuous and $C$ is connected, the sets $C^{\epsilon_1},\ldots, C^{\epsilon_k}$ are connected. Hence $C^{\epsilon_1}\cdots C^{\epsilon_k}$, the continuous image of the connected set $C^{\epsilon_1}\times \cdots \times C^{\epsilon_k}$ (in $\pi^k$) under multiplication, is connected. Since both $x$ and the identity belong to $C^{\epsilon_1}\cdots C^{\epsilon_k}$, it follows that $x$ belongs to the connected component of $\langle C\rangle$ at the identity. Since $x$ was arbitrary, $\langle C\rangle$ is connected.