Cyclic Group to Subgroup, Subring and Ideal Generated by a Subset

                             Cyclic Group to Subgroup, Subring and 

Ideal Generated by a Subset

M. Velrajan

The objective of this article is to develop mathematical thinking, problem solving and thereby developing critical thinking among young mathematics learners.

This can be done by enlightening the young minds with understanding and exploration of the beauties of pure mathematics and with developing interest in mathematics.

Let us discuss how we can find the subgroup generated by a subset of a group and the subring and ideal generated by a subset of a ring by understanding the cyclic subgroup of a group.

We start with 

1. Suppose * is an associative binary operation on a set A.

Then we know that * as a mapping from A × A into A, for any a, b ∈ A, a * b ∈ A and it is unique.

The natural way of thinking raises about 3 elements and in general, n elements. 

We note that, by induction on n, for any n elements a1, a2, . . . , an ∈ A, a1 * a2 * . . . * an ∈ A.

Of course note that the associativity of * is needed to define a1 * a2 * . . . * an , from a * b.

If the binary operation * is denoted by multiplication then for any n elements a1, a2, . . . , an ∈ A,

a1a2 . . . an ∈ A and in particular, for any a ∈ A, an ∈ A for all natural numbers n.

If the binary operation * is addition then for any n elements a1, a2, . . . , an ∈ A, 

a1 + a2 + . . . + an ∈ A and in particular, for any a ∈ A, na ∈ A for all natural numbers n.

This is a way of developing mathematical thinking. And we see how this observation is going to be the crux of our discussion.

  

2. Let G be a group and a ∈ G. 

Then what is the smallest subgroup of G containing a ? 

Let H be a subgroup of G containing a. Then a, a -1 ∈ H, and an, (a - 1)n = a - n ∈ H, for all n ∈ ℕ, by our observation in 1. Since a0 = e, we have an ∈ H, for all n ∈ ℤ. 

Note that if H is a subgroup then an ∈ H, for all n ∈ ℤ and a ∈ H.  

Hence {an / n ∈ ℤ} ⊆ H.

But {an / n ∈ ℤ} itself is a subgroup of G containing a

(since the set is clearly non-empty and an(am)-1 = an - m).

Hence {an / n ∈ ℤ} is the smallest subgroup of G containing a and it is called the cyclic subgroup generated by a and it is denoted by (a) or <a>.

2.1. Note that (a) = (a - 1) for all a ∈ G.

2.2. If the order of the element a is finite, say n, 

then <a> = {an / n ∈ ℤ} = {a0 = e, a, a2, . . . , an - 1} and it is isomorphic to (ℤn, ⊕n).

If the order of the element a is infinite then 

<a> = {an / n ∈ ℤ} = { . . . , a - 2, a - 1, a0 = e, a, a2, . . . , an, . . . } and it is isomorphic to (ℤ, +).

Understand how the subgroup definition and properties are used in the above proof.

Think of a generalization of the above.

3. Let G be a group and S be a nonempty subset of G. 

Then what is the smallest subgroup of G containing S ?

Let H be a subgroup of G containing S. 

Then a, a -1 ∈ H and an ∈ H, for all a ∈ S and for all n ∈ ℤ.

Aliter :

Let H be a subgroup of G containing S. Then x, x -1 ∈ H, for all x ∈ S. 

Then for all k ≥ 1 and x1, x2, . . . , xk ∈ S,

x1x2 . . . xk ∈ H, x1-1x2-1 . . . xk-1 ∈ H, x1-1x2x3x4-1x5 . . . xk-1 ∈ H, . . . . . 

In general, any finite product of either elements of S or inverse of elements of S are in H. 

Hence a1a2 . . . an ∈ H, where either ai ∈ S or ai-1 ∈ S for all i and n ∈ ℕ. 

i.e {a1a2 . . . an / either ai ∈ S or ai-1 ∈ S for all i and n ∈ ℕ} ⊆ H.

But {a1a2 . . . an / either ai ∈ S or ai-1 ∈ S for all i and n ∈ ℕ} itself is a subgroup of G containing S. (Since the above set is  clearly non-empty and if a1a2 . . . an, b1b2 . . . bm are two elements of the above set then 

a1a2 . . . an(b1b2 . . . bm)-1 = a1a2. . .anbm-1bm - 1-1. . . b1-1 is also a finite product of either elements of S or inverse of elements of S.)

Hence {a1a2 . . . an / either ai ∈ S or ai-1 ∈ S for all i and n ∈ ℕ} is the smallest subgroup of G containing S. 

3.3.1. If (G, +) is a group and S is a nonempty subset of G then 

(S) = {k1a1 + k2a2 + . . . + knan / a1, a2, . . . , an ∈ S and n ∈ ℕ, k1, k2, . . . , kn ∈ ℤ}.

3.3.2. If (G, +) is an abelian group and S is a nonempty subset of G then 

(S) = {k1a1 + k2a2 + . . . + knan / {a1, a2, . . . , an } ⊆ S and n ∈ ℕ, k1, k2, . . . , kn ∈ ℤ}.

We denote {k1a1 + k2a2 + . . . + knan / {a1, a2, . . . , an } ⊆ S and n ∈ ℕ, k1, k2, . . . , kn ∈ ℤ} by ℤS. 

Understand the above. Now think of the similar one to a ring.

4. Let R be a commutative ring with identity and S be a nonempty subset of R.

Then what is the smallest ideal of R containing S (or ideal <S> generated by S)?

If I is an ideal of R containing S then (I, +) is a subgroup of the abelian group (R, +). 

But ℤS = {k1a1 + k2a2 + . . . + knan / {a1, a2, . . . , an } ⊆ S and n ∈ ℕ, k1, k2, . . . , kn ∈ ℤ} is the smallest subgroup of (R, +) containing S. Hence ℤS ⊆ I. 

Also, for r ∈ R and x ∈ S, rx ∈ I. Since the sum of finite number of elements of I is in I, by 1,

for  r1, r2, . . . , rn ∈ R and  x1, x2, . . . , xn  ∈ S,  r1x1 + r2x2 +  . . . + rnxn ∈ I. 

i.e. {r1x1 + r2x2 + . . . + rnxn / n ∈ ℕ, r1, r2, . . . , rn ∈ R and  x1, x2, . . . , xn ∈ S} ⊆ I.

We denote {r1x1 + r2x2 + . . . + rnxn / n ∈ ℕ, r1, r2, . . . , rn ∈ R and  x1, x2, . . . , xn ∈ S} by RS. 

For all a ∈ S, and k ∈ ℤ, ka = a + a + . . . + a = 1a + 1a + . . . + 1a, if k > 0 and 

ka = - a - a - . . . - a = (-1)a + (-1)a + . . . + (-1)a, if k < 0. And ka = 0, if k = 0.

Since 1 ∈ R,  ka ∈ RS, for all a ∈ S and k ∈ ℤ.

Hence S ⊂ ℤS ⊆ RS ⊆ I.

But RS = {r1x1 + r2x2 + . . . + rnxn / n ∈ ℕ, r1, r2, . . . , rn ∈ R and  x1, x2, . . . , xn ∈ S} is an ideal of R(verify).

Hence RS = {r1x1 + r2x2 + . . . + rnxn / n ∈ ℕ, r1, r2, . . . , rn ∈ R and  x1, x2, . . . , xn ∈ S} is the smallest ideal of R containing S.

Note that since 1 ∈ R, we get that ℤS ⊆ RS. So what happens if R is without the identity.

5. If R is a commutative ring without the identity element  and S be a nonempty subset of R.

then what is the smallest ideal of R containing S?

If I is an ideal of R containing S then, as in 4, S ⊂ ℤS ⊆ I and RS ⊆ I. But we will not get that

ℤS ⊆ RS.  

But the smallest ideal containing S have to contain both ℤS and RS. Think of how to get a ideal containing both RS and ℤS and contained in I. 

So, we consider RS + ℤS.

Consider RS + ℤS = {r1x1 + . . . + rnxn + k1a1 + . . . + kmam / a1, . . . , am, x1, . . . , xn ∈ S, n, m ∈ ℕ,

                                                                                             k1, . . . , kn ∈ ℤ and r1, . . . , rn ∈ R} ⊆ I.  

Since RS and ℤS are subgroups of I, RS + ℤS is also a subgroup and  S ⊆ RS + ℤS ⊆ I.

For r ∈ R, x ∈ S, rx ∈ RS and hence nrx ∈ RS, for all n ∈ ℤ (since (RS, +) is a group). 

Also nrx = rx + rx + . . . + rx = r(x + x + . . . + x) = rnx, if n > 0,

nrx = - rx - rx - . . . - rx = r(- x - x - . . . - x) = rnx, if n < 0 and 0rx = 0 = r0 = r0x.

Hence rnx = nrx ∈ RS, for all r ∈ R, x ∈ S and n ∈ ℤ.  

Hence for r1x1 + . . . + rnxn + k1a1 + . . . + kmam ∈ RS + ℤS and r ∈ R,

r(r1x1 + . . . + rnxn + k1a1 + . . . + kmam) = rr1x1 + . . . + rrnxn + k1ra1 + . . . + kmram ∈ RS ⊆ RS + ℤS.

So, RS + ℤS is an ideal containing S.

Hence RS + ℤS = {r1x1 + . . . + rnxn + k1a1 + . . . + kmam / a1, . . . , am, x1, . . . , xn ∈ S, n, m ∈ ℕ, 

                                                                                                   k1, . . . , kn ∈ ℤ and r1, . . . , rn ∈ R}

 is the smallest ideal of R containing S.

6. If R is a ring without the identity element and S is a nonempty subset of R, what is the smallest ideal of R containing S ?

If S is a nonempty subset of a ring R without the identity element and I is an ideal of R containing S then, as in 5, 

RS + ℤS = {r1x1 + . . . + rnxn + k1a1 + . . . + kmam / a1, . . . , am, x1, . . . , xn ∈ S, n, m ∈ ℕ, 

                                                                                       k1, . . . , kn ∈ ℤ and r1, . . . , rn ∈ R} ⊆ I 

is a left ideal of R containing S.

Think how to get an ideal containing S and contained in I.

In any ideal J containing S, for all y ∈ S, ys ∈ J, for all y ∈ S and s ∈ R and hence  

y1s1 + . . . + ymsm ∈ J, for all yj ∈ S and sj ∈ R. Also tzu ∈ J, for all z ∈ S and t, u ∈ R and hence t1z1u1 + . . . + tpzpup ∈ J, for all zj ∈ S and tj, uj ∈ R. 

So, we consider RS + SR + RSR + ℤS 

Consider RS + SR + RSR + ℤS 

= {r1x1 + . . . + rnxn + y1s1 + . . . + ymsm + t1z1u1 + . . . + tpzpup + k1a1 + . . . +   kqaq / xi, yj, zk, al ∈ S, 

                                                                    n, m, p, q ∈ ℕ, k1, . . . , kq ∈ ℤ and ri, sj, tk, ul ∈ R} ⊆ I.

And it is an ideal containing S(verify). Hence RS + SR + RSR + ℤS =

{r1x1 + . . . + rnxn + y1s1 + . . . + ymsm + t1z1u1 + . . . + tpzpup + k1a1 + . . . + kqaq / xi, yj, zk, al ∈ S, 

                                                                           n, m, p, q ∈ ℕ, k1, . . . , kq ∈ ℤ and ri, sj, tk, ul ∈ R}

is the smallest ideal of R containing S.

6.1. If R is a ring with the identity element and S is a nonempty subset of R, then 

RS + SR + RSR ={r1x1 + . . . + rnxn + y1s1 + . . . + ymsm + t1z1u1 + . . . + tpzpup / xi, yj, zk ∈ S, 

                                                                                                        n, m, p ∈ ℕ, and ri, sj, tk, ul ∈ R}

is the smallest ideal of R containing S(since if 1 ∈ R then ℤS ⊆ RS).

7. If S = {x1, . . . , xn} is a finite subset of a ring R then the ideal <S> generated by S is called a finitely generated ideal of R and  

<S> = RS + SR + RSR + ℤS = {r1x1 + . . . + rnxn + x1s1 + . . . + xnsn + t1x1u1 + . . . + tnxnun + 

                                                                     k1x1 + . . . + knxn / k1, . . . , kn ∈ ℤ and ri, sj, tk, ul ∈ R}.

If R has identity and / or commutative write <S> suitably.

If a ∈ R then the ideal generated by {a} is called the principal ideal generated by a.

8. Let R be a ring and S be a nonempty subset of R.  

Then what is the smallest subring of R containing S ?

If T is a subring of R containing S then (T, +) is a subgroup of the abelian group (R, +),

ℤS = {k1a1 + k2a2 + . . . + knan / {a1, a2, . . . , an } ⊆ S and n ∈ ℕ, k1, k2, . . . , kn ∈ ℤ} is the smallest subgroup of (R, +) containing S. Hence ℤS ⊆ T.

Note that ℤS is not a subring of T. So, use the multiplication.

9. Examples

i) In the ring of integers (ℤ, +, .), for any n ∈ ℤ, the subgroup generated by n, 

ℤ{n} = {kn / k ∈ ℤ} = nℤ, which is also a subring of (ℤ, +, .). Also for all n ∈ ℤ, nℤ are ideals. 

i.e. ℤ{n} = ℤ[{n}] = nℤ = ideal generated by n. 

Since (ℤ, +) is cyclic and since subgroup of a cyclic group is cyclic, any subgroup of  (ℤ, +) is

ℤ{n} = nℤ, for some n ∈ ℤ. 

Hence nℤ, n ∈ ℤ, are the only subgroups of  (ℤ, +) and hence they are the only subrings and the only ideals of (ℤ, +, .).   

So, in (ℤ, +, .) all the subgroups are subrings and all the subrings are ideals.

ii) Let R be a ring with identity 1. Then the smallest subgroup of (R, +) containing {1} or the subgroup generated by 1 in (R, +) is ℤ{1} = {n1 / n ∈ ℤ}, which is also a subring of R. 

i.e. ℤ{1} = ℤ[{1}] = {n1 / n ∈ ℤ}. 

If the order of 1 in (R, +) is infinite then ℤ[{1}] = {n1 / n ∈ ℤ} is isomorphic to the ring of integers

(ℤ, +, .).

If the order of 1 in (R, +) is finite, say n, then  ℤ[{1}] = {k1 / k = 0, 1, 2, . . . , n - 1} is isomorphic to the ring of integers modulo n, (ℤn, ⊕n, ⊙n).

If the order of 1 in (R, +) is infinite then the order of 1+1 in (R, +) is infinite and hence 

ℤ{1+1} = {n(1+1) / n ∈ ℤ} is infinite. 

For n, m ∈ ℤ, (n(1+1))(m(1+1)) = nm(1+1)2 = nm(1+1+1+1) = 2nm(1+1). 

Hence ℤ{1+1} = {n(1+1) / n ∈ ℤ} is closed under multiplication and hence it is a subring of R.

Therefore, ℤ{1+1} = ℤ[{1+1}] = {n(1+1) / n ∈ ℤ} and it is isomorphic to the subring

(2ℤ, +, .) of (ℤ, +, .).

In general, for any integer k ≥ 1, the order of k1 in (R, +) is infinite and hence 

ℤ{k1} = {n(k1) / n ∈ ℤ}.

For n, m ∈ ℤ, (n(k1))(m(k1)) = nm(k1)2 = nm(k21) = knm(k1). 

Hence ℤ{k1} = {n(k1) / n ∈ ℤ} is closed under multiplication, hence it is a subring of R.

Therefore, ℤ{k1} = ℤ[{k1}] = {n(k1) / n ∈ ℤ} and it is isomorphic to the subring (kℤ, +, .) 

of (ℤ, +, .).

Note that for k < 0, ℤ{k1} = ℤ{(-k)1}.

  

10. Let V be a vector space over a field F and S be a nonempty subset of V. 

Then what is the smallest subspace of V containing S ?

If W is a subspace of V containing S then for 𝛂 ∈ F and x ∈ S, 𝛂x ∈ W and sum of finite number of elements of W is in W. 

Hence for 𝛂1, 𝛂2, . . . , 𝛂n ∈ F and  x1, x2, . . . , xn  ∈ S, 𝛂1x1 ⧾ 𝛂2x2 ⧾ . . . ⧾ 𝛂nxn ∈ W. 

But {𝛂1x1 ⧾ 𝛂2x2 ⧾ . . . ⧾ 𝛂nxn / n ∈ ℕ, 𝛂1, 𝛂2, . . . , 𝛂n ∈ F and  x1, x2, . . . , xn ∈ S} is a subspace of V containing S(verify). 

Hence {𝛂1x1 ⧾ 𝛂2x2 ⧾ . . . ⧾ 𝛂nxn / n ∈ ℕ, 𝛂1, 𝛂2, . . . , 𝛂n ∈ F and  x1, x2, . . . , xn ∈ S} is the smallest subspace of V containing S.

Any similar way of learning mathematics by understanding the concepts thoroughly, applying the learned concepts properly, thinking and imagining mathematically makes one love and enjoy mathematics. 

Enjoy Mathematics.