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One of the reasons we care about factor groups is that we can often determine information about large, possibly hard to explore groups, by taking information from the factor group and pulling it back to the larger group. We know that the map $f:G\to G/N$ defined by $f(g)=Ng$ is a homomorphism, so we can often obtain information about $G$ by using this homomorphism to pull back information from $G/N$. The next problem has you show an example of this, namely that if you can find an element of order $k$ in $G/N$, then there must be an element of this order as well in $G$. Your work from last time should make this problem quite fast, provided you remember that $\langle a\rangle$ is a cyclic group, and as such is isomorphic to $\mathbb{Z}_{|a|}$, which we understand quite well.
Problem 93 (If A Factor Group $G/N$ Has An Element Of Order $k$ Then So Does $G$)
Suppose that $N$ is a normal subgroup of a finite group $G$ and that $Na$ has order $k$ as an element of $G/N$. Prove the following two facts.
- The order of $a$ in $G$ is a multiple of the order of $Na$ in $G/N$.
- There exists an element $b\in G$ that has order $k=|Na|$.
In other words, if a factor group has an element of order $k$, the the original group before factoring must have an element of order $k$ as well.
How does normality behave when we look at subgroups? The following problem has you show that if you know $N$ is normal in $G$, and $N$ lies in some subgroup $B$ of $G$, then $N$ must be normal in $B$ as well. However, if we know $N$ is normal in $B$, and $B$ is normal in $G$, this does not guarantee that $N$ is normal in $G$. In symbols, we have $N\leq B\leq G$ and $N\trianglelefteq G$ implies $N\trianglelefteq B$, but we cannot say that $N\trianglelefteq B\trianglelefteq G$ implies $N\trianglelefteq G$. Being normal in a large group is enough to pass down to subgroups, but being normal in a subgroup is not enough to force normality in a larger group.
Problem 95 (Subgroups And Normality)
Suppose that $G$ is a group and that $A\leq B\leq G$.
- If $A$ is normal in $G$, prove that $A$ is normal in $B$.
- If $A$ is normal in $B$ and $B$ is normal in $G$, must $A$ be normal in $G$? Find a counter example to this.
The next problem has you show that when you compute factor groups, lots of properties pass immediately from $G$ to $G/N$. In particular, if $G$ is Abelian then $G/N$ is Abelian. Also, if $G$ is cyclic, then $G/N$ is cyclic. Many other properties follow as well.
Problem 96 (Factor Groups Preserve Being Cyclic And Abelian)
Suppose that $N$ is a normal subgroup of $G$.
- If $G$ is cyclic, prove that $G/N$ is cyclic.
- If $G$ is Abelian, prove that $G/N$ is Abelian.
The previous problem is an application of the bigger idea connected to homomorphism. Namely that the image of an Abelian group is Abelian, and the image of a cyclic group is cyclic.
Problem 97 (Images Of Abelian And Cyclic Groups)
Let $f:G\to H$ be a homomorphism. We have already shown that the image of $f$, written $f(G)$, is a subgroup of $H$.
- Prove that if $G$ is Abelian, then $f(G)$ is Abelian.
- Prove that if $G$ is cyclic, then $f(G)$ is cyclic.
How do the properties of Abelian and cyclic behave when combined with external direct products?
Problem 98 (External Direct Products Of Abelian And Cyclic Groups)
Suppose that $G$ and $H$ are groups.
- Prove or disprove: If both $G$ and $H$ are Abelian, then $G\oplus H$ is Abelian.
- Prove or disprove: If $G\oplus H$ is Abelian,then both $G$ and $H$ are Abelian.
- Prove or disprove: If both $G$ and $H$ are cyclic, then $G\oplus H$ is cyclic.
- Prove or disprove: If $G\oplus H$ is cyclic, then both $G$ and $H$ are cyclic.
The next problem has you prove Fermat's little lemma. We already conjectured this earlier in the semester when we noticed that if $p$ is a prime, then regardless of which $a$ we pick from $U(p)$, we must always have $a^{p-1}=1$. Sometimes, this is the order of $a$, whereas sometimes it is not. First, let's look at another conjecture we made earlier in the semester, to get us back into thinking about the $U$ groups.
Exercise (The Last Element Of UN Is Always Its Own Inverse)
Pick an integer $n$ greater than 1. Prove that $(n-1)\in U(n)$ is always its own multiplicative inverse.
Click to see a solution.
Notice that $(n-1)\pmod n = (-1)\pmod n$, which means $(n-1)^2\pmod n = (-1)^\pmod n = 1\pmod n$. This shows that the order of $n-1$ is 2, and also that $n-1$ is its own inverse.
Problem 99 (Fermat's Little Lemma)
Let $G$ be a finite group. Use Lagrange's theorem to prove the following two corollaries.
- We have $a^{|G|}=e$ for every $a\in G$.
- [Fermat's Little Lemma] Let $p$ be a prime and suppose $a\in U(p)$. Then we must have $a^p\pmod p=a$. (Hint: What is the order of $U(p)$?)
Using Fermat's little lemma and/or Lagrange's theorem, you should be able to rapidly prove the next result.
Exercise (The Order Of A In UP Divides P-1)
Let $p$ be a prime. Let $a\in U(p)$. Then $|a|$ divides $p-1$.
Click to see a solution.
There are several ways to prove this result.
- Since we know $a^p\pmod p = a$, we konw $a^{p-1}\pmod p = 1$. We know that $a^k=e$ in any group only when $k$ is a multiple of the order of $|a|$, so this means $k=p-1$ is a multiple of $|a|$.
- We now that $|U(p)|=p-1$. Hence every subgroup of $U(n)$, in particular $\left<a\right>$, must have an order that divides $p-1$. Since we know $|a|=|\left<a\right>|$, we know that $|a|$ divides $p-1$.
Because we are working with homomorphisms quite a bit in our work, we see the notation $f(g)=g$ 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. Next time we'll show 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 $G$.
For more problems, see AllProblems