Prove or disprove that $H$ is a subgroup of $C^*$ under multiplication.

Asked Mar 23 '17 at 11:49

Active Mar 23 '17 at 11:58

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Prove or disprove that $H$ is a subgroup of $C^*$ under multiplication.

Let $$H = \{a + bi \,|\, a, b R, a^2 + b^2=1\}$$ I know for every $a \in H$, if we can prove its inverse is in $H$ then we are done.

I said let $a+bi$ and $c+di$ belong to $H$. Then $$(a + bi)(c + di)^{-1} = (ac + bd) + (bc - ad)i$$ and $$(ac + bd)^2 + (bc - ad)^2 = 1$$

Hence $H$ is a subgroup of $C^*$ under multiplication.

Yes? No? How can I be more detailed in my answer? I do have to teach this back to someone. Thank you.

edited Mar 23 '17 at 11:58

lioness99a

asked Mar 23 '17 at 11:49

M.Matthews

https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Mar 23 '17 at 11:53
This seems confused in a few ways. It's not true that given a subset $S$ of a group $G$ that if $S$ is closed under the inverse map then $S$ is a subgroup. For any group $G$ and any nonidentity element $g \in G$, $S = {g, g^{-1}}$ does not contain the identity and hence is not a subgroup. – Travis Willse Mar 23 '17 at 11:54

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