Group Isomorphisms and Homomorphisms

We have gone through many isomorphisms, so I will provide this as just an example and test.

$U(7)$ is isomorphic to ${\mathbb{Z}}_6$

Could be useful to take a look at Behaviors of Z modulo Firstly note: $U(7)=\{1,2,3,4,5,6\}$ ${\mathbb{Z}}_6=\{0,1,2,3,4,5\}$ First is a group of integers under multiplication modulo 7, and second is a group of integers under addition modulo 6.

First lets prove an isomorphism from $\varphi :{\mathbb{Z}}_6\to U(7)$. Notice, 3 generates $U(7)$ and 1 generates ${\mathbb{Z}}_6$. Thus from here we can start noticing that any isomorphism must take $\varphi (1)=3$. Thus, for any element $n$ in ${\mathbb{Z}}_6$ we have $\varphi (n)=\varphi (1_1+1_2+...+1_n)=\varphi (1{)}_1\star \varphi (1{)}_2\star ...\star \varphi (1{)}_n=3^n$. From here we can clearly just define the isomorphism as $\varphi (n)=3^n$ and clearly this is a homomorphism as: $\varphi (n+m)=3^{n+m}=3^n{3}^m=\varphi (n)\varphi (m)$ and its bijective as ${\varphi }^{-1}(3^n)=lo{g}_3(3^n)$

$\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic

1. For contradiction assume they are isomorphic, and we know that 1 generates all of $\mathbb{Z}$. Then there must be an element s.t. $\varphi (1)=a$. And then, that element generate all of $\mathbb{Q}$. We can notice $\mathbb{Q}$ is not cyclic and stop here, but we can also notice that there must be an element which maps to $\frac{a}{2}$. So we have it such that $\varphi (x)=\frac{a}{2}$ and $\varphi (x+x)=\varphi (x)+\varphi (x)=\frac{a}{2}+\frac{a}{2}=a=\varphi (1)$
2. as $\frac{1}{2}$ does not exist in $\mathbb{Z}$ we can conclude there is no such element and this is not an isomorphism.

Important Classifications:

1. Any cyclic group is either isomorphic to $\mathbb{Z}$ if infinite or ${\mathbb{Z}}_n$ if finite
2. Any group with prime order is cyclic, so any group with p elements is isomorphic to ${\mathbb{Z}}_p$.

Proof that any group with prime order is cyclic:

1. Let $G$ be a group with order prime $p$. Then we have it so any element $a\in G$ must generate a subgroup of $G$. However, by langrange’s theorem the order of this group must divide the order of $G$. As $G$ is order $p$ is has no devisors, thus a must generate the whole group and therefore the group is cyclic.

Cayley’s Theorem

Any group is isomorphic to a group of permutations. You can take a table and define a permutation for each element. Ill skip this for now as its extremely geometric but the book provides a good explanation.

Direct Products

Product of two groups, $G_1\times G_2$ creates another group where we define the operation as $(g_1,g_1^{‘})(g_2,g_2^{‘})=(g_1{g}_2^{‘},g_1^{‘}{g}_2)$ And the order is the LCM of the order of both groups, or $LCM(\mid G_1\mid ,\mid G_2\mid )$. Its like 2 wheels turning and the first time they both become 0 is the LCM of their cycles. Similar to problem 2.1 in Practice Midterm 1

Internal Direct Products of ${\mathbb{Z}}_n$ with relatively prime orders are isomorphic to their product.

${\mathbb{Z}}_n\times {\mathbb{Z}}_m={\mathbb{Z}}_{mn}⟺GCD(n,m)=1$ This is because their GCD being 1 implies they have no common devisors, so their LCM is their product ($nm$) which means they generate the entire group before reaching their product, making it a cyclic group with $nm$ elements and isomorphic to ${\mathbb{Z}}_{mn}$

I.E: Take ${\mathbb{Z}}_3$ and ${\mathbb{Z}}_5$. Then, as $GCD(3,5)=1$ we know this is isomorphic to ${\mathbb{Z}}_{15}$.

You can do this in reverse and take the prime factorization of something and show its isomorphic: $60=2\ast 2\ast 3\ast 5$ so then we have it that ${\mathbb{Z}}_{60}={\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_3\times {\mathbb{Z}}_5$.