Question:

True/False?

Every infinite abelian group has atleast one element of infinite order.

Hint: Characteristic of a ring.

Discussion: What if we have an element of finite order, and attach it to infinitely many “inert” elements?

Consider \(\mathbb{Z_2}\). ( Note that, \(\mathbb{Z_n}\) is an integral domain iff n is prime. And a finite integral domain is a field. Therefore, \(\mathbb{Z_2}\) is a (finite) field. )

\(\mathbb{Z_2}[x]\) is the ring of polynomials with coefficients from \(\mathbb{Z_2}\). A ring is a group with some extra conditions. \(\mathbb{Z_2}[x]\) is a group with respect to addition. What is the order of each element in this group?

Let \(p(x)=a_0+a_1x+a_2x^2+…a_nx^n\in\mathbb{Z_2}[x]\)

Since \(a_0,a_1,…,a_n\in\mathbb{Z_2}\) we have \(2a_0,2a_1,…,2a_n=0\). Note that 0 here means the identity element of the group \(\mathbb{Z_2})\), not the 0 we use in Real number system.

Hence, \(p(x)+p(x)=2p(x)=0\). Here 0 is the identity of \(\mathbb{Z_2}[x]\).

Therefore, each element of the group \((\mathbb{Z_2}[x],+)\) has order 1 or 2.

The elements \((1,x,x^2,x^3,…)\) are all distinct members of this group, hence \((\mathbb{Z_2}[x],+)\) is an infinite group, where every element has finite order.

Isomorphic examples: The set of all sequence with elements from \(\mathbb{Z_2}\) & the infinite Cartesian product \(\mathbb{Z_2}\times \mathbb{Z_2}\times\mathbb{Z_2}\times…\) are two examples which follows.

Non-isomorphic simple generalizations: Of course, there is nothing special about the number 2. We can take any prime (natural) number. Can we take any natural number? Although in courses of algebra we usually deal with \(\mathbb{Z_p}\), p-prime, here there is no need to restrict ourselves to prime p-s only. Indeed we can take any \(\mathbb{Z_n}[x]\), with \(n\in\mathbb{N}\). It may not be ‘good enough’ as a ring (\(\mathbb{Z_n}\) not being integral domain for n not prime and all that…) but it is sufficient for our example.

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