I think I answered this correctly, but I wanted to be sure.Determine whether the given subset H of the group G is a subgroup.

Problem:

$\displaystyle G = \mathbb{C}^{x}$, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.

$\displaystyle H = \left\{z \in \mathbb{C} \hspace{2mm} |\hspace{2mm} |z| = 1\right\}$

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For H to be a subgroup of G, the following three things must be satisfied:

1) Closed under the binary operation of G (in this case, multiplication of complex numbers)

2) H has the identity element of G

3) For all $\displaystyle a \in H$, there exists an $\displaystyle a^{-1} \in H$.

$\displaystyle |z|$ is defined as $\displaystyle |z|=\sqrt{a^2+b^2}$ whereaandbcome from the general complex number form $\displaystyle a+bi$.

Since $\displaystyle |z| = 1$, it must be true that only one of the pairaandbis 1, and the other needs to be zero.

Therefore, this set H only has four elements:

$\displaystyle 1+0i$

$\displaystyle 0+1i$

$\displaystyle -1+0i$

$\displaystyle 0-1i$

1) Multiplying two elements of this set will always return a result that is still within the set H, and therefore the set H isclosed under the binary operation. Since there are always going to be two zero terms when foiling two of these terms, only the terms where +/-1 are multiplied together, always returning one of the four elements in the group.

2) An identity of group $\displaystyle G$ will satisfy

$\displaystyle a*e = e*a = a$

In this case, $\displaystyle e=1+0i$ intuitively satisfies this, and is the unique identity element of group G. $\displaystyle |e|=|1+0i|=1$, so the set H has the identity of G.

3) The inverse of a complex number is the conjugate. So a complex number $\displaystyle x = a+bi$ has the inverse $\displaystyle x^{-1}=a-bi$. Looking at the list of four elements from part (1), it can be seen that the first two elements are $\displaystyle x$ and $\displaystyle y$, and the following two are the conjugates (inverses) of $\displaystyle x$ and $\displaystyle y$, and vice versa. Therefore, every element in H has an inverse in H.

Since (1), (2), and (3) are satisfied, H is a subgroup of G.

Am I correct? Is there anything here that I did wrong?