20131125

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The next problem has you look at a very specific homomorphism from $G\times H$ to $G$. It's the start of a key idea that shows $(G\times H)/H\approx G$, namely that if you first create an external product, and then decide later to annihilate one of your factors, then you will obtain the other factor. It's basically an extension of the division fact from integers that says $(xy)/y=x$. A similar result shows that $(G\times H)/G\approx H$.

Problem (Collapsing A Factor Of An External Direct Product Yields The Other Factor, or $(G\times H)/H\approx G$)

Let $G$ and $H$ be groups, and consider the projection map $\pi_1:G\times H\to G$ defined by $\pi_1(g,h)=g$. We call this the projection map $\pi_1$ because it takes the element $(g,h)$ and returns the first component. The projection map $\pi_2$ would return the second component $h$.

1. Prove that this projection map is a surjective homomorphism.
2. What is the kernel of $\pi_1$, in other words what is $f^{-1}(e_G)$?
3. Why is $(G\times H)/(\{e_G\}\times H)\approx G$?
4. Show that $i_H:\{e_G\}\times H\to H$ defined by $i_H(e_G,h)=h$ is an isomorphism.

It's time to start using the power of factor groups, together with induction, to start proving some powerful theorems that apply to any finite group. The key to many of the problems we'll see in the next 4 weeks 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 first of these theorems is often called Cauchy's theorem. 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 $p$. The theorem below has you show this fact is true for Abelian groups. 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 we will eventually remove.

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 $p$.

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 $\varphi(15\cdot 63)$.

Problem (Introduction To Internal Direct Products)

Consider the group $G\times H$. Let $A=G\times \{e_H\}$ and $B=\{e_G\}\times H$. Prove the following.

1. $A\cap B$ contains only the identity element of $G\times H$.
2. The set product $AB$ equals the whole group $G\times H$.
3. $A$ and $B$ are normal subgroups of $G\times H$.
4. $G\approx A$ and $H\approx B$, which means $G\times H=AB\cong A\times B$.

Did you notice what we did above. We started with a group created by using an external direct product. We then found two subgroup $A$ and $B$ of the group, and showed that the group was equal to the set product $AB$ (a product done internally in the group) and at the same time isomorphic to the external direct product $A\times B$. The first three conditions are the key.

Definition (Internal Direct Product Of Two Subgroups)

Suppose that $G$ is a group. If $H$ and $K$ are normal subgroups of $G$ such that $H\cap K=\{e\}$ and $HK=G$, then we say that $G$ is the internal direct product of $H$ and $K$.

Problem (When Is $HK$ A Subgroup Of $G$)

Suppose $H$ and $K$ are subgroups of $G$.

1. Prove that $HK$ is a subgroup of $G$ if and only if $HK=KH$.
2. Prove that if either $H$ or $K$ is normal in $G$, then $HK$ is a subgroup of $G$.

For more problems, see AllProblems