Please Login to access more options.

---

Today

« December 2016 »
Sun	Mon	Tue	Wed	Thu	Fri	Sat
				1	2	3
4	5	6	7	8	9	10
11	12	13	14	15	16	17
18	19	20	21	22	23	24
25	26	27	28	29	30	31

---

Because we are working with homomorphisms quite a bit in our work, we see the notation $f(g)=h$ quite a bit. We have also seen the notation $f^{-1}(h)$, but because homomomorphisms are not necessarily injective, this is not the inverse of $f$ applied to $h$. We have to remember that when we write $f^{-1}(h)$ we are talking about a set, namely the set of all $x\in G$ such that $f(x)=h$. We are not talking about a single $g$ so that $f(g)=h$. The notation $f^{-1}(h)$ represents the preimage (or inverse image) of $\{h\}$ under $f$. Here's a reminder of the formal definition.

Definition (Preimage Or Inverse Image)

If $f:A\to B$ and $C\subseteq B$, then the preimage of $C$ under $f$ is the set of all $a\in A$ such that $f(a)\in C$, and we write the preimage of $C$ under $f$ as $$f^{-1}(C)=\{a\in A\mid f(a)\in C\}.$$ When $C=\{c\}$ contains a single element, we often write the preimage of $C$ under $f$ as $f^{-1}(c)$ instead of using the more formal cumbersome notation $f^{-1}(\{c\})$.

---

Preimages and subgroups interact in a very nice way, namely homomorphisms preserve subgroups when computing inverse images. It is also true that preimages preserve normality.

Problem 100 (The Preimage Of A Subgroup Under A Homomorphism Is A Subgroup)

Suppose that $f:G\to H$ is a homomorphism. Let $B$ be a subgroup of $H$. Let $A$ be the preimage of $B$, namely $$A=f^{-1}(B)=\{g\in G\mid f(g)\in B\}.$$ Show that $A$ is a subgroup of $G$.

---

Problem 101 (Homomorphisms Preserve Normal Subgroups)

Suppose that $f:G\to H$ is a homomorphism. Use The Normal Subgroup Test to prove the following:

1. If $N$ is normal in $G$, then $f(N)$ is normal in $f(G)$.
2. If $B$ is normal in $H$, then $f^{-1}(B)$ is normal in $G$.

---

At this point, we've proved quite a large collection of facts about homomorphisms and what they preserve. I strongly suggest that you look at pages 202-204 of your text to see a list of these properties. Then read the remarks on page 204.

The following two facts follow immediately from using the properties of homomorphisms. The first fact shows that any time you have a subgroup that contains half the elements of the group, that subgroup must be a normal subgroup.

Problem 102 (Subgroups Of Index 2 Are Normal)

Suppose that $H$ is a subgroup of $G$ with index $|G:H|=2$. Recall that the index of $H$ in $G$ is the number of distinct cosets of $H$ in $G$.

1. Build a surjective homomorphism from $G$ to $\mathbb{Z}_2$.
2. Show that $H$ is normal in $G$.

---

This next fact shows that any time you have subgroup in a factor group $G/N$, it corresponds to a subgroup $H$ in $G$ that contains $N$.

Problem 103 (Subgroups Of A Quotient Group Correspond To Subgroups Of The Original Group)

Suppose that $N$ is a normal subgroup of $G$ and that $B$ is a subgroup of $G/N$. Prove the following:

1. There exists a subgroup $H\leq G$ such that $H/N=B$.
2. If we know that $n=|N|$ and $m=|B|$ (so $n$ is the order of $N$ in $G$, and $m$ is the order of $B$ in $G/N$), then $G$ must have a subgroup $H$ of order $nm$.

As a suggestion, consider the homomorphism $f:G\to G/N$ given by $f(g)=Ng$ and use some properties of homomorphisms.

---

To tackle the next problem, you'll want to rely on the fact that $|(g,h)|=\text{lcm}(|g|,|h|)$, together with the fact that two integers are relatively prime if and only if their product is their least common multiple.

Problem 104 (A Direct Product Of Cyclic Groups Is Cyclic If And Only If The Groups Have Relatively Prime Orders)

Suppose that $G$ and $H$ are finite cyclic groups. Prove that $G\oplus H$ is cyclic if and only if $|G|$ and $|H|$ are relatively prime.

Exercise (A Direct Product Of Cyclic Groups Is Cyclic If And Only If The Groups Have Relatively Prime Orders)

Use induction to extend the previous result to the external direct product of $n$ cyclic groups. So prove that if $G_1,G_2,\ldots,G_n$ are $n$ cyclic groups, then $G_1\oplus G_2\oplus \cdots\oplus G_n$ is cyclic if and only if $|G_i|$ and $|G_j|$ are relatively prime when $i\neq j$.

Problem 105 (An Isomorphism From $U(st)$ To $U(s)\oplus U(t)$ When $s$ And $t$ Are Relatively Prime)

Recall that $\varphi$, the Euler-phi function is defined by $\varphi(n)=|U(n)|$, the order of $U(n)$. Suppose $n=st$ where $s$ and $t$ are relatively prime. In this problem you'll show that $\varphi(st)=\varphi(s)\varphi(t)$.

1. Show that $f:U(st)\to U(s)\oplus U(t)$ defined by $f(x)=(x\mod s,x\mod t)$ is an isomorphism.
2. Why does $\varphi(st)=\varphi(s)\varphi(t)$ when $s$ and $t$ are relatively prime?
3. Use the results above to compute $\varphi(17\cdot 19)$ and $\varphi(15\cdot 63)$.

---

It's time to use the power of factor groups, together with induction, to prove the first of many powerful theorems that apply to finite groups. The key lies in the fact that the order of a factor group $G/N$ will always be less than the order of $G$, provided $N$ is not trivial. This means that if we use induction on the order of a group, then if we can use a factor group to pull back information to a larger group, induction will provide us with precisely the tool we need to prove some really powerful theorems about groups.

The next theorem is often called Cauchy's theorem (for Abelian groups). The more general version of Cauchy's theorem states that if $p$ is a prime that divides the order of the group, then $G$ must have an element of order $p$. Because it's an Abelian group, every subgroup is normal, which means we can quickly create factor groups any time we obtain a subgroup. If we remove the Abelian condition, then we have to prove that a subgroup is normal before we can create a factor group. The Abelian condition is a simplifying assumption, which can be removed with much more work.

Problem 106 (Abelian Groups Have An Element Of Order $p$ For Every Prime That Divides The Order Of The Group)

Let $G$ be an Abelian group and let $p$ be a prime. Prove that if $p$ divides the order of $G$, then there exists $a\in G$ with order $p$.

---

---

For more problems, see AllProblems