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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.