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I know this question has been asked before, but I would like to specifically address one part of the proof I'm reading that confuses me. I am following Fraleigh 7th edition, page 213.

The theorem is stated as follows:

If $G$ is a finite subgroup of the multiplicative group $\langle F^*, \cdot\rangle$ of a field $F$, then $G$ is cyclic.

Proof:
By theorem 11.12 as a finite abelian group, $G$ is isomorphic to a direct product $Z_{d_1}\times Z_{d_2} \times \dots \times Z_{d_r}$, where each $d_i$ is a power of a prime. Let us think of each of the $Z_{d_i}$ as a cyclic group of order $d_i$ in multiplicative notation. Let $m$ be the least common multiple of all the $d_i$ for $i = 1, 2, \dots, r$; note that $m \le d_1d_2\dots d_r$. [Here's where I get lost] If $a_i$ is in $Z_{d_t}$, then $a_i^{d_i} = 1$. The proof continues from there, but this is the only spot in the proof that confuses me.

To me, if you take any element $a_i$ in $Z_{d_t}$ and compute $a_i^{d_i}$, you get $0$, not $1$. In fact, it's complete nonsense that $a_i^{d_i}$ is always $1$. For example, $2^4$ in $Z_4$ is $16 \bmod 4 = 0$.

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  • $\begingroup$ $\mathbb{Z}_4$ isn't a field; it has zero divisors. $\endgroup$
    – user61527
    May 12 '14 at 22:29
  • $\begingroup$ Guide to MathJax $\endgroup$
    – user61527
    May 12 '14 at 22:30
  • $\begingroup$ I feel like I followed the guide exactly. $\endgroup$
    – user7348
    May 12 '14 at 22:33
  • $\begingroup$ You have to insert dollar signs to render it as math. $\endgroup$
    – user61527
    May 12 '14 at 22:33
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    $\begingroup$ The text says the groups $Z_{d_i}$ you're using are multiplicative. The notation isn't really good, I'd have used $C_{d_i}$ to mean a multiplicative cyclic group of order $d_i$. $\endgroup$
    – egreg
    May 12 '14 at 22:38
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$G$ is isomorphic to a product of cyclic groups $\mathbb{Z}/d_1\mathbb{Z}\times \dots \times\mathbb{Z}/d_r\mathbb{Z} $ and here each element $a\in \mathbb{Z}/d_1\mathbb{Z}$ satisfies $(d_1)a=0$, as the group law is additive.

But $G$ is a subgroup of the multiplicative group of a field, so the group law on it is the multiplication. This is why you can see the element $a$ in $G$ and say $a^{d_1}=1$. It is just a question of writing groups additively or multiplicatively.

In fact, the isomorphism $\varphi\colon G\to \mathbb{Z}/d_1\mathbb{Z}\times \dots \times\mathbb{Z}/d_r\mathbb{Z}$ satisfies $\varphi(gh)=\varphi(g)+\varphi(h)$ for each $g,h\in G$, since the group law at the source is multiplicative and the group law at the target is additive.

You can also see $Z_{d_i}$ as $Z_{d_i}=\langle x\mid x^{d_i}=1\rangle$ and see these multiplicatively.

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  • $\begingroup$ So, we're simply viewing 1 in G as 0 in the target under the isomorphism? e.g. Since, phi(1) = phi(1*1) = phi(1) + phi(1) which implies phi(1) = 0. So, we're free to talk of 0 as 1? $\endgroup$
    – user7348
    May 12 '14 at 23:08
  • $\begingroup$ Yes, in any group there is an element called identity, which satisfies that if you apply the group law of it with $g$, then you get $g$. If you write the group multiplicatively, then you usually write the identity $1$ and if you write the group additively, the identity is written $0$. And the isomorphism $\varphi$ sends the identity of $G$ (written $1$) on the identity of $\mathbb{Z}/d_1\mathbb{Z}\times \dots \times\mathbb{Z}/d_r\mathbb{Z}$, which is $(0,\dots,0)$. $\endgroup$ May 13 '14 at 7:25

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