## A Book of Abstract Algebra, Second Edition (1982)

### Chapter 16. THE FUNDAMENTAL HOMOMORPHISM THEOREM

Let *G* be any group. In __Chapter 15__ we saw that every quotient group of *G* is a homomorphic image of *G*. Now we will see that, conversely, every homomorphic image of *G* is a quotient group of *G*. More exactly, *every homomorphic image of G is isomorphic to a quotient group of G*.

It will follow that, for any groups *G* and *H*, *H* is a homomorphic image of *G* iff *H* is (or is isomorphic to) a quotient group of *G*. Therefore, the notions of *homomorphic image* and of *quotient group* are interchangeable.

The thread of our reasoning begins with a simple theorem.

**Theorem 1** *Let f*: *G* → *H be a homomorphism with kernel K*. *Then*

*f*(*a*) = *f*(*b*)*iffKa* = *Kb*

(In other words, any two elements *a* and *b* in *G* have the *same image* under *f* iff they are in the same coset of *K*.)

Indeed,

What does this theorem really tell us? It says that if *f* is a homomorphism from *G* to *H* with kernel *K*, then all the elements in any fixed coset of *K* have the same image, and, conversely, elements which have the same image are in the same coset of *K*.

It is therefore clear already that there is a one-to-one correspondence matching cosets of *K* with elements in *H*. It remains only to show that this correspondence is an isomorphism. But first, how exactly does this correspondence match up specific cosets of *K* with specific elements of *H*? Clearly, for each *x*, the coset *Kx* is matched with the element *f*(*x*). Once this is understood, the next theorem is easy.

**Theorem 2** *Let f: G* → *H* *be a homomorphism of* *G* onto *H*. If *K is the kernel of f, then*

*H* ≅ *G/K*

PROOF: To show that *G*/*K* is isomorphic to *H*, we must look for an isomorphism from *G*/*K* to *H*. We have just seen that there is a function from *G*/*K* to *H* which matches each coset *Kx* with the element *f*(*x*); call this function *ϕ*. Thus, *ϕ* is defined by the identity

*ϕ*(*Kx*) = *f*(*x*)

This definition does not make it obvious that *ϕ*(*Kx*) is *uniquely defined*. (If it is not, then we cannot properly call *ϕ* a function.) We must make sure that if *Ka* is the same coset as *Kb*, then *ϕ*(*Ka*) is the same as *ϕ*(*Kb*): that is,

*if* *Ka* = *Kb* *then* *f*(*a*) = *f*(*b*)

As a matter of fact, this is true by __Theorem 1__.

Now, let us show that *ϕ* is an isomorphism:

*ϕ is injective:* If *ϕ*(*Ka*) = *ϕ*(*Kb* then *f*(*a*) = *f*(*b*); so by __Theorem 1__, *Ka* = *Kb*.

*ϕ is surjective*, because every element of *H* is of the form *f*(*x*) = *ϕ*(*Kx*). Finally, *ϕ*(*Ka · Kb*) = *ϕ*(*Kab*) = *f*(*ab*) = *f*(*a*)*f*(*b*) = *ϕ*(*Ka*)*ϕ*(*Kb*).

Thus, *ϕ* is an isomorphism from *G*/*K* onto *H*. ■

__Theorem 2__ is often called the *fundamental homomorphism theorem*. It asserts that every homomorphic image of *G* is isomorphic to a quotient group of G. Which specific quotient group of *G*? Well, if *f* is a homomorphism from *G* onto *H*, then *H* is isomorphic to the quotient group of *G* by the kernel of *f*

The fact that *f* is a homomorphism from *G* *onto H* may be symbolized by writing

Furthermore, the fact that *K* is the kernel of this homomorphism may be indicated by writing

Thus, in capsule form, the fundamental homomorphism theorem says that

Let us see a few examples:

We saw in the opening paragraph of __Chapter 14__ that

is a homomorphism from _{6} onto _{3}. Visibly, the kernel of *f* is {0, 3}, which is the subgroup of _{6} generated by ⟨3⟩, that is, the subgroup (3). This situation may be symbolized by writing

We conclude by __Theorem 2__ that

_{3} ≅ _{6}/⟨3⟩

For another kind of example, let *G* and *H* be any groups and consider their direct product *G* × *H*.Remember that *G* × *H* consists of all the ordered pairs (*x*, *y*) as *x* ranges over *G* and *y* ranges over *H*. You multiply ordered pairs by multiplying corresponding components; that is, the operation on *G* × *H* is given by

(*a*, *b*) · (*c*, *d*) = (*ac*, *bd*)

Now, let *f* be the function from *G* × *H* onto *H* given by

*f*(*x*, *y*) = *y*

It is easy to check that *f* is a homomorphism. Furthermore, (*x*, *y*) is in the kernel of *f* iff *f* (*x*, *y*) = *y* = *e*. This means that the kernel of *f* consists of all the ordered pairs whose second component is *e*. Call this kernel *G**; then

G* = {(*x*, *e*): *x* ∈ *G*}

We symbolize all this by writing

By the fundamental homomorphism theorem, we deduce that *H* ≅ (*G* × *H*)/*G**.[It is easy to see that *G** is an isomorphic copy of *G*; thus, identifying *G** with *G*, we have shown that, roughly speaking, (*G* × *H*)/*G* ≅ *H*.]

Other uses of the fundamental homomorphism theorem are given in the exercises.

**EXERCISES**

In the exercises which follow, FHT will be used as an abbreviation for fundamental homomorphism theorem.

**A. Examples of the FHT Applied to Finite Groups**

In each of the following, use the fundamental homomorphism theorem to prove that the two given groups are isomorphic. Then display their tables.

**Example** _{2} *and* _{6}/⟨2⟩.

is a homomorphism from _{6} onto _{2}.(Do not prove that *f* is a homomorphism.) The kernel of *f* is {0, 2, 4} = ⟨2⟩. Thus,

It follows by the FHT that _{2} ≅ _{6}/⟨2⟩.

**1** _{5} and _{20}/⟨5⟩.

**2** _{3} and _{6}/⟨3⟩.

**3** _{2} and *S*_{3}/{*ε*, *β*, *δ*}.

**4** *P*_{2} and *P*_{3}/*K*, where *K* = {0, {*c*}}. [HINT: Consider the function *f*(*C*) = *C* ∩ {*a*, *b*}. *P*_{3} is the group of subsets of {*a*, *b*, *c*}, and *P*_{2} of {*a*, *b*}.]

**5** _{3} and (_{3} × _{3})/*K*, where *K* = {(0, 0), (1, 1), (2, 2)}.[HINT: Consider the function *f*(*a*, *b*) = *a* − *b* from _{3} × _{3} to _{3}.]

**B. Example of the FHT Applied to ()**

Let *α*: () → be defined by *α*(*f*) = *f*(l) and let *β*:() → be defined by *β*(*f*) = *f*(2).

**1** Prove that *α* and *β* are homomorphisms from () *onto* .

**2** Let *J* be the set of all the functions from to whose graph passes through the point (1, 0) and let *K* be the set of all the functions whose graph passes through (2, 0).Use the FHT to prove that ≅ ()/*J* and ≅ ()/*K*

**3** Conclude that ()/*J* ≅ ()/*K*

**C. Example of the FHT Applied to Abelian Groups**

Let *G* be an abelian group. Let *H* = {*x*^{2}: *x* ∈ *G*} and *K* = {*x ∈ G*: *x*^{2} = *e*}.

**1** Prove that *f*(*x*) = *x*^{2} is a homomorphism of *G onto H*.

**2** Find the kernel of *f*.

**3** Use the FHT to conclude that *H* ≅ *G*/*K*

**† D. Group of Inner Automorphisms of a Group G**

Let *G* be a group. By an *automorphism* of *G* we mean an isomorphism *f*: *G* → *G*.

# ** 1** The symbol Aut(

*G*) is used to designate the set of all the automorphisms of

*G*.Prove that the set Aut (

*G*), with the operation ∘ of composition, is a group

*by proving that Aut*(

*G*)

*is a subgroup of S*.

_{G}**2** By an *inner automorphism* of *G* we mean any function *ϕ _{a}* of the following form:

for every *x* ∈ *Gϕ _{a}*(

*x*) =

*axa*

^{−}

^{1}

Prove that every inner automorphism of *G* is an automorphism of *G*.

**3** Prove that, for arbitrary *a*, *b* ∈ *G*.

*ϕ _{a}* ∘

*ϕ*=

_{b}*ϕ*

_{ab}and(

*ϕ*)

_{a}^{−}

^{1}=

*ϕ*

_{a}_{ }

_{−}

_{ 1}

**4** Let *I*(*G*) designate the set of all the inner automorphisms of *G*. That is, *I*(*G*) = {*ϕ _{a}*:

*a*∈

*G*}.Use part 3 to prove that

*I*(

*G*) is a subgroup of Aut(

*G*).Explain why

*I*(

*G*) is a group.

**5** By the *center* of *G* we mean the set of all those elements of *G* which commute with every element of *G*, that is, the set *C* defined by

*C* = {*a* ∈ *G*: *ax* = *xa* for every *x* ∈ *G*}

Prove that *a* ∈ *C* if and only if *axa*^{−}^{1} = *x* for every *x* ∈ *G*.

**6** Let *h*: *G* → *I*(*G*) be the function defined by *h(a*) = *ϕ _{a}*. Prove that

*h*is a homomorphism from

*G*onto

*I*(

*G*) and that

*C*is its kernel.

**7** Use the FHT to conclude that *I*(*G*) is isomorphic with *G*/*C*.

**† E.The FHT Applied to Direct Products of Groups**

Let *G* and *H* be groups. Suppose *J* is a normal subgroup of *G* and *K* is a normal subgroup of *H*.

**1** Show that the function *f*(*x*, *y*) = (*Jx*, *Ky*) is a homomorphism from *G* × *H onto* (*G*/*J*) × (*H*/*K*).

**2** Find the kernel of *f*.

**3** Use the FHT to conclude that (*G* × *H*)/(*J* ×K) ≅ (*G*/*J*) × (*H*/*K*).

**† F. First Isomorphism Theorem**

Let *G* be a group; let *H* and *K* be subgroups of *G*, with *H* a normal subgroup of *G*. Prove the following:

**1** *H* *K* is a normal subgroup of *K*

# ** 2** If

*HK*= {

*xy*:

*x*∈

*H*and

*y*∈

*K*}, then

*HK*is a subgroup of

*G*.

**3** *H* is a normal subgroup of *HK*.

**4** Every member of the quotient group *HK*/*H* may be written in the form *Hk* for some *k* ∈ *K*.

**5** The function *f*(*k*) = *Hk* is a homomorphism from *K* onto *HK*/*H*, and its kernel is *H* *K*

**6** By the FHT, *K*/(*H* *K*) ≅ *HK*/*H*. (This is referred to as the *first isomorphism theorem*.)

**† G. A Sharper Cayley Theorem**

If *H* is a subgroup of a group *G*, let *X* designate the set of all the left cosets of *H* in *G*. For each element *a* ∈ *G*, define *p _{a}*:

*X*→

*X*as follows:

*p _{a}*(

*xH*) = (

*ax*)

*H*

**1** Prove that each *p _{a}* is a permutation of

*X*.

**2** Prove that *h*: *G* → *S _{X}* defined by

*h*(

*a*) =

*p*is a homomorphism.

_{a}# ** 3** Prove that the set {

*a*∈

*H*:

*xax*

^{−}

^{1}∈

*H*for every

*x*∈

*G*}, that is, the set of all the elements of

*H*whose conjugates are all in

*H*, is the kernel of

*h*.

**4** Prove that if *H* contains no normal subgroup of *G* except {*e*}, then *G* is isomorphic to a subgroup of *S _{X}*.

**† H. Quotient Groups Isomorphic to the Circle Group**

Every complex number *a* + *b***i** may be represented as a point in the complex plane.

The *unit circle* in the complex plane consists of all the complex numbers whose distance from the origin is 1; thus, clearly, the unit circle consists of all the complex numbers which can be written in the form

cos *x* + **i** sin *x*

for some real number *x*.

# ** 1** For each

*x*∈ , it is conventional to write cis

*x*= cos

*x*+ i sin

*x*. Prove that eis (

*x*+

*y*) = (cis

*x*)(cis

*y*).

**2** Let *T* designate the set {cis *x*: *x* ∈ }, that is, the set of all the complex numbers lying on the unit circle, with the operation of multiplication. Use part 1 to prove that *T* is a group. (*T* is called the *circle group*.)

**3** Prove that *f*(*x*) = cis *x* is a homomorphism from onto *T*.

**4** Prove that ker *f* = {2*nπ*: *n* ∈ } = ⟨2*π*⟩.

**5** Use the FHT to conclude that *T* ≅ /⟨2⟩.

**6** Prove that *g*(*x*) = cis 2*πx* is a homomorphism from onto *T*, with kernel .

**7** Conclude that *T* ≅ /.

**† I. The Second Isomorphism Theorem**

Let *H* and *K* be normal subgroups of a group *G*, with *H* *k* Define *ϕ*: *G*/*H* → *G*/*K* by *ϕ*(*Ha*) = *Ka*.

Prove parts 1–4:

**1** ϕ is a well-defined function. [That is, if *Ha* = *Hb*, then *ϕ*(*Ha*) = *ϕ*(*Hb*).]

**2** *ϕ* is a homomorphism.

**3** *ϕ* is surjective.

**4** ker *ϕ* *K*/*H*

**5** Conclude (using the FHT) that (*G*/*H*)*K*/*H*) ≅ *G /K*.

**† J. The Correspondence Theorem**

Let *f* be a homomorphism from G onto *H* with kernel *K*:

If *S* is any subgroup of *H*, let *S** = {*x ∈G*: *f*(*x*)∈ *S*}. Prove:

**1** *S** is a subgroup of *G*.

**2** *K* *S**.

**3** Let *g* be the restriction of *f* to *S*.*[That is, *g*(*x*) = *f*(*x*) for every *x ∈ S**, and *S** is the domain of *g*.] Then *g* is a homomorphism from *S** onto *S*, and *K* = ker g.

**4** *S* ≅ *S**/*K*.

**† K. Cauchy’s Theorem**

*Prerequisites*: * Chapter 13, Exercise I, and Chapter 15, Exercises G and H*.

If *G* is a group and *p* is any prime divisor of |*G*|, it will be shown here that *G* has at least one element of order *p*. This has already been shown for abelian groups in __Chapter 15__, __Exercise H4__. Thus, assume here that *G* is not abelian. The argument will proceed by induction; thus, let |*G*| = *k*, and assume our claim is true for any group of order less than *k*. Let C be the center of *G*, let *C _{a}* be the centralizer of

*a*for each

*a*∈

*G*, and let

*k*=

*c*+

*k*+ ⋯ +

_{s}*k*be the class equation of

_{t}*G*, as in

__Chapter 15__,

__Exercise G2__.

**1** Prove: If *p* is a factor of |*C _{a}*| for any

*a*∈

*G*, where

*a*∉

**C**, we are done. (Explain why.)

**2** Prove that for any *a* ∉ **C** in *G*, if *p* is not a factor of |*C _{a}*|, then

*p*is a factor of (

*G*:

*C*).

_{a}**3** Solving the equation *k* = *c* + *k _{s}* + ⋯ +

*k*for

_{t}*c*, explain why

*p*is a factor of

*c*. We are now done. (Explain why.)

**† L. Subgroups of p-Groups (Prelude to Sylow)**

*Prerequisites*: * Exercise J*;

__Chapter 15__,

__Exercises G__and

__H__.

Let pbea prime number. A *p-group* is any group whose order is a power of *p*. It will be shown here that if |*G*| = *p ^{k}* then

*G*has a normal subgroup of order

*p*for every

^{m}*m*between 1 and

*k*. The proof is by induction on |

*G*|; we therefore assume our result is true for all /^-groups smaller than

*G*. Prove parts 1 and 2:

**1** There is an element *a* in the center of *G* such that ord *(a*) = *p*. (See __Chapter 15__, __Exercises G__ and __H__.)

**2** ⟨*a*⟩ is a normal subgroup of *G*.

**3** Explain why it may be assumed that *G*/⟨*a*⟩ has a normal subgroup of order *p ^{m}*

^{ }

^{−}

^{1}.

# ** 4** Use

__Exercise J4__to prove that

*G*has a normal subgroup of order

*p*.

^{m}**SUPPLEMENTARY EXERCISES**

Exercise sets M through Q are included as a challenge for the ambitious reader. Two important results of group theory are proved in these exercises: one is called Sylow’s theorem, the other is called the basis theorem of finite abelian groups.

**† M. p-Sylow Subgroups**

*Prerequisites: Exercises J and K of this Chapter, Exercise* I1 of

__Chapter 14__, and

__Exercise D3__of

__Chapter 15__.

Let *p* be a prime number. A finite group *G* is called a *p-group* if the order of every element *x* in *G* is a power *p*. (The orders of different elements may be different powers of *p*.) If *H* is a subgroup of any finite group *G*, and *H* is a *p*-group, we call *H* a *p-subgroip* of *G*. Finally, if *K* is a *p*-subgroup of *G*, and *K* is maximal (in the sense that *K* is not contained in any larger *p*-subgroup of *G*), then *K* is called a *p*-*Sylow subgroup* of *G*.

**1** Prove that the order of any *p*-group is a power of *p*.(HINT:Use __Exercise K__.)

**2** Prove that every conjugate of a *p*-Sylow subgroup of *G* is a *p*-Sylow subgroup of *G*.

Let *K* be a *p*-Sylow subgroup of *G*, and *N* = *N*(*K*) the normalizer of *K*.

**3** Let *a* ∈ *N*, and suppose the order of *Ka* in *N*/*K* is a power of *p*. Let *S* = ⟨*Ka*⟩ be the cyclic subgroup of *N*/*K* generated by *Ka*. Prove that *N* has a subgroup *S** such that *S*/K* is a *p*-group. (HINT: See __Exercise J4__.)

**4** Prove that *S** is a *p*-subgroup of *G* (use __Exercise D3__, __Chapter 15__). Then explain why *S** = *K*, and why it follows that *Ka* = *K*.

**5** Use parts 3 and 4 to prove: no element of *N*/*K* has order a power of *p* (except, trivially, the identity element).

**6** If *a* ∈ *N* and the order of *p* is a power of *p*, then the order of *Ka* (in *N*/*K*) is also a power of *p*. (Why?) Thus, *Ka* = *K*.(Why?)

**7** Use part 6 to prove: if *aKa*^{−}^{l} = *K* and the order of *a* is a power of *p*, then *a* ∈ *K*.

**† N. Sylow’s Theorem**

*Prerequisites*: * Exercises K* and

*of this Chapter and*

__M____Exercise I__of

__Chapter 14__.

Let *G* be a finite group, and *K* a *p*-Sylow subgroup of *G*.Let *X* be the set of all the conjugates of *K*. See __Exercise M2__. If *C*_{1}, *C*_{2} ∈ *X*, let *C*_{1}∼*C*_{2} iff *C*_{1} = *aC*_{2}*a ^{−}^{l}* for some

*α*∈

*K*

**1** Prove that ∼ is an equivalence relation on *X*.

Thus, ∼ partitions *X* into equivalence classes. If *C* ∈, *X* let the equivalence class of *C* be denoted by [C].

**2** For each *C* ∈ *X*, prove that the number of elements in [*C*] is a divisor of *|K|*. (HINT: Use __Exercise I10__ of __Chapter 14__.) Conclude that for each *C* ∈ *X*, the number of elements in [*C*] is either 1 or a power of *p*.

**3** Use __Exercise M7__ to prove that the only class with a single element is [*K*],

**4** Use parts 2 and 3 to prove that the number of elements in *X* is *kp* + 1, for some integer *k*.

**5** Use part 4 to prove that (*G:N*) is *not* a multiple of *p*.

**6** Prove that (*N: K*) is *not* a multiple of *p*. (Use __Exercises K__ and __M5__.)

**7** Use parts 5 and 6 to prove that (*G*: *K*) is *not* a multiple of *p*.

**8** Conclude: Let *G* be a finite group of order *p ^{k}m*, where

*p*is not a factor of

*m*. Every

*p*-Sylow subgroup

*K of G*has order

*p*.

^{k}Combining part 8 with __Exercise L__ gives

*Let G be a finite group and let p be a prime number*. *For each n such that p ^{n} divides* |

*G*|,

*G*

*has a subgroup of order p*.

^{n}This is known as Sylow’s theorem.

**† O. Lifting Elements from Cosets**

The purpose of this exercise is to prove a property of cosets which is needed in __Exercise Q__. Let *G* be a finite abelian group, and let *a* be an element of *G* such that ord(*a*) is a multiple of ord(*x*) for every *x* ∈ *G*. Let *H* = ⟨*a*⟩. We will prove:

*For every x* ∈ *G*, *there is some* *y* ∈ *G* *such that Hx* = *Hy and* ord(*y*) = *ord*(*Hy*).

This means that every coset of *H* contains an element *y* whose order is the same as the coset’s order.

Let *x* be any element in *G*, and let ord *(a*) = *t*, ord*(x*) = *s*, and ord (*Hx*) = *r*.

**1** Explain why *r* is the least positive integer such that *x ^{r}* equals some power of

*a*, say

*x*=

^{r}*a*.

^{m}**2** Deduce from our hypotheses that *r* divides *s*, and *s* divides *t*.

Thus, we may write *s* = *ru* and *t* = *sυ*, so in particular, *t* = *ruυ*.

**3** Explain why *a ^{mu}* =

*e*, and why it follows that

*mu*=

*tz*for some integer

*z*. Then explain why

*m*=

*rυz*.

**4** Setting *y* = *xa*^{−}* ^{υz}*, prove that

*Hx*=

*Hy*and ord(

*y*) =

*r*, as required.

**† P. Decomposition of a Finite Abelian Group into p-Groups**

Let *G* be an abelian group of order *p ^{k}m*, where

*p*and

^{k}*m*are relatively prime (that is,

*p*and

^{k}*m*have no common factors except ±1). (REMARK: If two integers

*j*and

*k*are relatively prime, then there are integers

*s*and

*t*such that

*sj*+

*tk*= 1. This is proved on page 220.)

Let *G _{p}^{k}* be the subgroup of

*G*consisting of all elements whose order divides

*p*. Let

^{k}*G*be the subgroup of

_{m}*G*consisting of all elements whose order divides ra. Prove:

**1** For any *x* ∈ *G* and integers *s* and *t*, *x ^{sp}^{k}* ∈

*G*and

_{m}*x*∈

^{tm}*G*.

_{p}^{k}**2** For every *x* ∈ *G*, there are *y* ∈ *G _{p}^{k}* and

*z*∈

*G*such that

_{m}*x*=

*yz*.

**3** *G _{p}^{k}*

*G*= {

_{m}*e*}.

**4** *G* ≅ *G _{p}^{k}* ×

*G*. (See

_{m}__Exercise H__,

__Chapter 14__.)

**5** Suppose |*G*| has the following factorization into primes: . Then *G* ≅ *G*_{1} × *G*_{2} × ⋯ × *G _{n}* where for each

*i*= 1, …,

*n*,

*G*is a

_{i}*p*-group.

_{i}**Q. Basis Theorem for Finite Abelian Groups**

*Prerequisite: Exercise P*.

As a provisional definition, let us call a finite abelian group “decomposable” if there are elements *a*_{l}, …, *a _{n}* ∈

*G*such that:

(Dl) For every *x* ∈ *G*, there are integers *k*_{1}, …, *k _{n}* such that . (D2) If there are integers

*l*

_{1}, …,

*l*such that then .

_{n}If (Dl) and (D2) hold, we will write *G* = [*a*_{1}, *a*_{2}, …, *a _{n}*]. Assume this in parts 1 and 2.

**1** Let *G*′ be the set of all products range over . Prove that *G*′ is a subgroup of *G*, and *G*′ = [*a*_{2}, …, *a _{n}*].

**2** Prove: *G* ≅ ⟨*a*_{1}⟩ × *G*′. Conclude that *G* ≅⟨*a*_{1}⟩ × ⟨*a*_{2}⟩ × ⋯ × ⟨*a _{n}*

In the remaining exercises of this set, let *p* be a prime number, and assume *G* is a finite abelian group such that the order of every element in *G* is some power of *p*. Let *a* ∈ *G* be an element whose order is the highest possible in *G*. We will argue by induction to prove that *G* is “decomposable.” Let *H* = ⟨*a*⟩.

**3** Explain why we may assume that *G*/*H* = [*Hb*_{1}, …, *Hb _{n}*] for some

*b*

_{l}, …,

*b*∈

_{n}*G*.

By __Exercise O__, we may assume that for each *i* = 1, …, *n*, ord (*b _{i}*) = (

*Hb*). We will show that

_{i}*G*= [

*a*,

*b*

_{1}, …,

*b*].

_{n}**4** Prove that for every *x* ∈ *G*, there are integers *k*_{0}, *k*_{1}, …, *k _{n}* such that

**5** Prove that if , then . Conclude that *G* = [*a*, *b*,_{1}, …, *b _{n}*].

**6** Use __Exercise P5__, together with parts 2 and 5 above, to prove: Every finite abelian group G is a direct product of cyclic groups of prime power order. (This is called the basis theorem of finite abelian groups.)

It can be proved that the above decomposition of a finite abelian group into cyclic *p*-groups is unique, except for the order of the factors. We leave it to the ambitious reader to supply the proof of uniqueness.