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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!

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    $\begingroup$ try to find an element $x$ and an integer $n$ such that $x^n = 1$ in each of the groups.(of course such that $x\neq 1$) $\endgroup$ – Thomas Sep 1 '17 at 16:12
  • $\begingroup$ Have a look at how I edited your question so that you learn some LaTeX on the way. $\endgroup$ – Ittay Weiss Sep 1 '17 at 16:18
  • $\begingroup$ @Thomas, you could post that hint as a hint-answer :-) $\endgroup$ – Jyrki Lahtonen Sep 1 '17 at 16:21
  • $\begingroup$ Thank you, I know LaTeX i didnt know you could put it on here. Ill have to look into how on here. Here you use 1 as the identity correct? And the exponential notation is just repeated group operation? $\endgroup$ – Alex Van de Kleut Sep 1 '17 at 16:24
  • $\begingroup$ How could I go about thinking of properties to test when testing for isomorphism? $\endgroup$ – Alex Van de Kleut Sep 1 '17 at 16:24
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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.

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    $\begingroup$ There are more elements of finite order, like $\frac 1{\sqrt 2}(1+I)$, but you have enough. $\endgroup$ – Ross Millikan Sep 1 '17 at 17:34
  • $\begingroup$ Yes, you are right. Thanks. $\endgroup$ – Sachchidanand Prasad Sep 1 '17 at 18:17
  • $\begingroup$ I haven't been exposed to concepts like order yet. Only the concept of a group being a set with a binary operation, associativity of the binary operation, an identity, and the existence of inverses for every element. $\endgroup$ – Alex Van de Kleut Sep 1 '17 at 19:08
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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.

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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.

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