Proving Finite subgroups of the multiplicative group of a field are cyclic

[Deanmark]

I am looking at this proof and I am stuck on the logic that $a^{p}$ = 1. For example, consider the group under multiplication without zero, ${Z}_{5}$, wouldn't 2^4 = 1 imply that the order is 4 not 5? We know that if G is a finite abelian group, G is isomorphic to a direct product ℤ(p1)^n1×ℤ(p2)^n2×⋯×ℤ(pr)^nr where pi's are prime not necessarily distinct. Consider each of the ℤ(pi)ni as a cyclic group of oreder pnii in multiplicative notation. Let m be the lcm of all the pnii for i=1,2,…,r. Clearly m≤p1n1p2n2⋯prnr. If ai∈ℤ(pi)ni then (ai)(pini)=1 and hence ami=1. Therefore for all α∈G, whe have αm=1 that is every element of G is a root of xm=1. However G has p1n1p2n2⋯prnr elements, while the polynomial xm−1 can have at most m roots in F. So, we deduce that m=p1n1p2n2⋯prnr. Therefore pi's are distinct primes, and the group G is isomorphic to the cyclic group ℤm.

 

[castor28]

I am looking at this proof and I am stuck on the logic that $a^{p}$ = 1. For example, consider the group under multiplication without zero, ${Z}_{5}$, wouldn't 2^4 = 1 imply that the order is 4 not 5?

Hi Deanmark,

The additive subgroup of $\mathbb{Z}_5$ has indeed order $5$, but the multiplicative subgroup only contains the non-zero elements, and has order $4$.

 

[Deanmark]

Hi Deanmark,

The additive subgroup of $\mathbb{Z}_5$ has indeed order $5$, but the multiplicative subgroup only contains the non-zero elements, and has order $4$.

Isn't $2^0$ the first element and $2^4$ the fifth operation which results in the identity making ${\Bbb{Z}}_{5}$ order 5? I guess this is what the proof is referring to when it claims for x in ${Z}_{d}$ $x^d = 1 $ where d is a prime power.

 

[castor28]

Isn't $2^0$ the first element and $2^4$ the fifth operation which results in the identity making ${\Bbb{Z}}_{5}$ order 5? I guess this is what the proof is referring to when it claims for x in ${Z}_{d}$ $x^d = 1 $ where d is a prime power.

Hi,

In $\mathbb{Z}_5$, $2^4 = 2^0 = 1$ (this is Fermat's theorem): the group only contains $4$ elements.

The $p_i$ in the proof are the prime divisors of the order of the multiplicative group under consideration. In the case of $\mathbb{Z}_5$, the complete multiplicative group has order $4$ (there is also a subgroup of order $2$). The theorem says that the group is isomorphic to $C_4$, not $C_2\times C_2$, because, otherwise, the equation $x^2 = 1$ would have four roots, which is impossible in a field.

The order of the multiplicative group (4 in this case) should not be confused with the order of the field (5). For example, the theorem also applies to proper subgroups (which is nothing new, since any subgroup of a cyclic group is cyclic). The theorem also applies in the case of a finite multiplicative subgroup of an infinite field: in $\mathbb{C}$, the n-th roots of unity constitute a finite cyclic multiplicative group.