Confusing theorem on cyclic groups and abelian groups

Question

There are 2 theorems above and can I seek some clarity on them? Is $C_8$ isomorphic to $C_2 \times C_2 \times C_2$?

From Theorem 4.23, it is not because the prime are not distinct From Theorem 5.16, it is since cyclic groups are abelian.

Answer 1

Is $C_8$ isomorphic to $C_2 \times C_2 \times C_2$?

No. The simple argument here is that $C_8$ has element of order $8$ while $C_2 \times C_2 \times C_2$ does not.

From Theorem 4.23, it is not because the prime are not distinct

Incorrect. Theorem 4.23 does not imply these are not isomorphic. This theorem only says what is isomorphic, it doesn't say what isn't isomorphic.

Ultimately it is like you say, but it does not follow from Theorem 4.23.

From Theorem 5.16, it is since cyclic groups are abelian.

Incorrect as well. The only thing you can conclude from Theorem 5.16 (together with observations about number of elements) is that $C_8$ is isomorphic to one of those three groups: $C_8$, $C_2\times C_4$, $C_2\times C_2\times C_2$.

The fact that these three groups are not pairwise isomorphic is not covered by those two theorems. In particular you need another theorem here:

Theorem. Let $p$ be prime. Then $$C_{p^{a_1}}\times\cdots\times C_{p^{a_n}}\simeq C_{p^{b_1}}\times\cdots\times C_{p^{b_m}}$$ if and only if $n=m$ and there exists reordering such that $a_i=b_i$ for all $i$.

And it is not hard to prove. I already gave a hint earlier: you need to look at elements of highest order. And then you can conclude something from that.

In particular these theorems together show that there are only three (up to isomorphism) abelian groups of order $8$: $C_8$, $C_2\times C_4$ and $C_2\times C_2\times C_2$. And only $C_8$ is cyclic among them.