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s1 = {e^i2πt ∈ C |t ∈ [0, 1]} Show that S1 is a subgroup of (C \ {0}, ×). I know there has to be an identity element and i have solved that as subbing in 0 for t gives you one.But i'm struggling to prove if g ∈ s1 then g^-1 ∈ s1 and also if g,h ∈ s1 then gh ∈ s1

I've proved it is a homomorphism also and need to use this to prove that (R/Z, +) ∼= (S1, ×).

2 Answers 2

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Hint:

$$g=e^{2\pi it}\;,\;\;0\le t\le 1\implies g^{-1}=e^{-2\pi it}=e^{2\pi i(1-t)}\;,\;\;0\le1-t\le 1$$

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    So for h would i just call t something different like use k and prove it using k? – 2017-02-17
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    @Pkr96 What? What I showed is that the inverse of an element in $\;S^1\;$ is also an element of $\;S^1\;$ by your definition... – 2017-02-17
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    Yeah i get that i was asking about g*h being an element of s1 – 2017-02-17
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    @Pkr96 Just like above: $$e^{2\pi it}\cdot e^{2\pi ik}=e^{2\pi i(t+k)}$$ and if $\;t+k\le1\;$ we're done, **otherwise** $$t+k=1+r\;,\;\;0\le r\le 1\implies e^{2\pi i(t+k)}=e^{2\pi ir}$$ and we're done. – 2017-02-17
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In complement, here is another way of proving that $\mathbb{S}^1$ is a subgroup of $(\mathbb{C} \smallsetminus \lbrace 0 \rbrace, \times)$. If $G$ and $G'$ are two groups and $f \, : \, G \, \rightarrow \, G'$ a group homomorphism, then:

  • $\mathrm{Im}(f) = \lbrace f(g), \, g \in G \rbrace$ is a subgroup of $G'$,
  • $\mathrm{ker}(f) = \lbrace g \in G, \, f(g) = 1_{G'} \rbrace$ is a subgroup of $G$.

Here, consider :

$G = (\mathbb{R},+)$, $G'=(\mathbb{C} \smallsetminus \lbrace 0 \rbrace, \times)$ and $f \, : \, G \, \rightarrow \, G'$ such that : $\forall t \in \mathbb{R}, \, f(t) = \exp(2i\pi t)$.