Show unit circle under multiplication not isomorphic to either the nonzero real numbers under multiplication or real numbers under addition I am trying to answer Fraleigh's Abstract Algebra textbook question from Exercises 4 (pg 45) q. 9.
"Show that the group $(U,\cdot)$ is NOT isomorphic to either $(\mathbb R^*,\cdot)$ or $(\mathbb R,+)$. All three groups have cardinality $|\mathbb R|$."
Here, $U$ is the group $\{z \in \mathbb C \mid  |z| = 1\}$.
I have no idea of the answer. I cant find any properties that are unique. I can show that the second two groups arent isomorphic but I have no idea how to begin.
I'd like if someone can give me a hint on how to approach. Thanks!
 A: In the group $U$ there  are more than two elements  whose order is finite namely, $i,-i,1,-1$. But in other two groups, it's not true. As in $(\mathbb{R}^*,\times)$ only $\pm1$ are of finite order and in $(\mathbb{R},+)$ there is no element of finite order.
A: As you haven't got order concepts of an element right now, I am giving my answer in elaborating concept of order concept.
Two groups are isomorphic if they share every group theoretic property. That is they are only notationally different.
Now note that the group $(U,.)$ has four elements $1, -1, i, -i$ having the property $x^4=1$. So if $(\mathbb R^*,.)$ is isomorphic to $(U,.)$ it should contain distinct four element with similar property. But $(\mathbb R^*,.)$ have only two elements namely $1$ and $-1$ such that $x^4=1.$ 
Also if we consider the the group $(\mathbb R,+)$ then it only have one element $0$ satisfying the property $4x=0$(as operation is $+,$ equation is written in addition format). So they are not isomorphic as we can't say groups are only notationally different as they are different in property also.
A: The group $U$ contains elements $x,y$ such that $x^3=y^3$ but $x\ne y.$ Neither of the other two groups has the corresponding property.
