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I'm a very beginner in Algebra trying to understand an isomorphism example.

Consider complex multiplication on the unit circle U where $|z| =1$ and addition modulo $2 \pi$ on $\Bbb R_{2\pi} := [ 0,2 \pi)$.

Fraleigh states:

$$\text{if}\quad z_1 \leftrightarrow \theta_1\quad\text{and}\quad z_2 \leftrightarrow\theta_2,\quad\text{then}\quad z_1*z_2 \leftrightarrow (\theta_1 +_{2\pi} \theta_2).$$ $$isomorphism$$

Here $ z\leftrightarrow\theta$ denotes "the natural one-to-one correspondence" between $z\in U$ and $\theta\in\Bbb R_{2\pi}$ defined by $z=e^{i\theta}$.

I understand that $z_i$ and $\theta_i$ have a one-to-one correspondence since for each $z_i$ there is exactly one $\theta_i$, thus if we know the value of $\theta_i$ we can directly determine $z_i$

However this is not the case for $z_1*z_2\leftrightarrow\theta_1 +_{2\pi} \theta_2$ , since the value of $\theta_1 +_{2\pi} \theta_2$ can be paired with many $z_i*z_j$. Suppose $\theta_1 +_{2pi} \theta_2 = \pi$ then both $e^{i \pi}*e^{i 0} $ and $e^{i \frac{\pi}{2}}*e^{i \frac{\pi}{2}}\leftrightarrow\pi$, hence the value of $\theta_1 +_{2\pi} \theta_2$ can be paired with both products. This is not a one-to-one correspondence

Reference: Fraleigh, A First Course in Abstract Algebra, Part 1 , page 16.

Revelant quote: "Two sets $X$ and $Y$ have the same cardinality if there exists a one-to-one function mapping $X$ onto $Y$, that is if there exists a one-to-one correspondence between $X$ and $Y$."

The author denotes a one-to-one correspondence as "$\leftrightarrow$" which is some sort of bijection.

In this case we have pairs ($z_i*z_j$, $\theta_i +_{2 \pi} \theta_j$) which we could regard as a function/mapping but as shown above it seems the value of right-side can be paired with many different products.

Anne Bauval
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qubitz
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  • This is not about the isomorphism but the one-to-one correspondance – qubitz Aug 03 '24 at 23:54
  • I added a relevant quote, note that the author does not define isomorphism prior this example and only tries to give a meaning to it through that example. I am more confused about the one-to-one correspondance between the product and the sum above. – qubitz Aug 04 '24 at 00:05
  • That is what he wrote exactly, I am attaching a picture with the text – qubitz Aug 04 '24 at 00:21
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    The "one-to-one correspondence" is not "between the product and the sum above". It is a bijection $f:[0,2\pi)\to U$ (namely: $f(t)=e^{it}$). What the author means is just that this $f$ has the following nice property: for every $s,t\in[0,2\pi)$, letting $z_1:=f(s),z_2:=f(t)$, we have $z_1z_2=f(s+_{2\pi}t)$. Nothing more. – Anne Bauval Aug 04 '24 at 00:23
  • I think I was/am confused because if the product $z_i * z_j = z_k* z_l $ then we view them as the same thing so there will still be a one-to-one correspondance if we refer to the values only is this correct? hence why the example with (pi,0) and (pi/2, pi/2) still yields a one-to-one correspondance even tho i != k and j != l , it's all good as long as i+j = l+k correct ? I was too focused on the notation – qubitz Aug 04 '24 at 00:57
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    "there will still be a one-to-one correspondance if we refer to the values only" doesn't make sense to me. A one-to-one correspondence is just a bijection between two sets (here: $[0,2\pi)$ and $U$). – Anne Bauval Aug 04 '24 at 05:35
  • Same question here. – Jean Marie Aug 04 '24 at 08:14
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    Related, but really not the same. The OP's problem is much more basic. – Anne Bauval Aug 04 '24 at 08:19
  • I'm sorry, @qubitz; it was around 6am here when I sent that comment. I'm not usually up that early. – Shaun Aug 04 '24 at 10:46
  • Did you mean to write $\Bbb R_{2\pi}$ instead of $\Bbb R_{2pi}$? – José Carlos Santos Aug 04 '24 at 11:06

2 Answers2

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then both $e^{i \pi}*e^{i 0} $ and $e^{i \frac{\pi}{2}}*e^{i \frac{\pi}{2}}\leftrightarrow\pi$, hence the value of $\theta_1 +_{2\pi} \theta_2$ can be paired with both products

They're the same product, the same single element of $U$.

no comment
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I think you're a bit confused here. It's not true that every $\theta_1+_{2\pi}\theta_2$ can correspond to more than one $z_1*z_2$; what is true is that it can correspond to more than one pair $(z_1,z_2)$ for which $z_1*z_2 \leftrightarrow \theta_1+_{2\pi}\theta_2$. But that's not relevant here for the bijection to hold.

In the example you gave, both $e^{i0}*e^{i\pi}$ and $e^{i\pi/2}*e^{i\pi/2}$ are $e^{i\pi}$. So they're the same $z_1*z_2$, even though the $z_1$ and $z_2$ are different.

The bijection is just saying that $e^{i\theta}$ when multiplying complex numbers, corresponds to $\theta$ when adding mod $2\pi$. It's really simple, you're probably overcomplicating it in your mind!

YiFan Tey
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