Math 441 - Abstract Algebra - Example - ExamplesOfTheFirstIsomorphismTheorem

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The first isomorphism theorem states

if $f:G\to H$ is a surjective homomorphism with kernel $K$, then the map $\phi:G/K\to H$ defined by $\phi(Kg)=f(g)$ is an isomorphism.

We've been using this since the first weeks of the semester. Let's examine what this means in the context modular arithmetic.

The group isomorphism $\mathbb{Z}/17\mathbb{Z}\approx \mathbb{Z}_{17}$

Let $G=\mathbb{Z}$ and $H = \mathbb{Z}_{17}$, and use the map $f: G\to H$ defined by $f(x) = x\mod 17$. Note that the kernel of $f$ is precise the multiples of 17, so we have $K=17\mathbb{Z}$. The first isomorphism theorem then says that $\mathbb{Z}/17\mathbb{Z}\approx \mathbb{Z}_{17}$. What is the difference?

In $\mathbb{Z}_{17}$, the only elements are the integers from 0 to 16. No other integers exists, and all discussion about elements must refer to only these 17 elements. Each time we perform an operation, the mod operation is performed.
In $\mathbb{Z}/17\mathbb{Z}$, we look at cosets. We know that $14+17\mathbb{Z} = -3+17\mathbb{Z}$ because $14-(-3)\in17\mathbb{Z}$. We use coset properties to perform computations and have no problem working with integers that are negative, or greater than 16. The mod operation is never performed, rather all elements that have the same remainder are just collected together in a single coset. No modding is required, rather we are free to pick the most convenient way to express the element for future computations.

Let's look at these differences in terms of a computation. We will compute $14+(14+(14+(14+14)))) = 5(14)$.

Performing the operations exclusively in $\mathbb{Z}_{17}$, we obtain $$ \begin{align} 14+(14+(14+(14+14)))) &= 14+(14+(14+11))) &&(28\mod 17 = 11)\\ &= 14+(14+8)) &&(25\mod 17 = 18)\\ &= 14+5 &&(22\mod 17 = 5)\\ &= 2 &&(19\mod 17 = 2). \end{align}$$ At each step above, we had to compute a sum, and then simplify mod 17.
Performing the operations in $\mathbb{Z}/17\mathbb{Z}$ allows us to freely choose a convenient element in the coset. Since we know $14 + 17\mathbb{Z} = -3+17\mathbb{Z}$, we obtain $$ \begin{align} 5(14+17\mathbb{Z})&=14+ 17\mathbb{Z}+(14+ 17\mathbb{Z}+(14+ 17\mathbb{Z}+(14+ 17\mathbb{Z}+14+ 17\mathbb{Z}))))\\ &=-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+14+ 17\mathbb{Z}))))\\ &=-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(11+ 17\mathbb{Z})))\\ &=-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(8+ 17\mathbb{Z}))\\ &=-3+ 17\mathbb{Z}+(5+ 17\mathbb{Z})\\ &=2+ 17\mathbb{Z}. \end{align}$$ The $+17\mathbb{Z}$ that appears on everything above is tedious to write, so often we just simply write

$$5(14)=-3-3-3-3+14 = -3-3-3+11 =-3-3+8 =-3+5 = 2,$$ and then note that this gives the set equality $5(14+17\mathbb{Z}) = 2+17\mathbb{Z}$.

The ring isomorphism $\mathbb{Z}/17\mathbb{Z}\approx \mathbb{Z}_{17}$

The computations above use the single binary operation + for both groups. When we add the binary operation $\cdot$ to the mix, we get ring structures. The first isomorphism theorem also holds for rings. This allows us to compute the product above much faster. We first note that $14+(14+(14+(14+14)))) = 5(14)$, so we can now use both multiplication and addition.

Performing the operations exclusively in the ring $\mathbb{Z}_{17}$, we obtain $$ \begin{align} 14+(14+(14+(14+14)))) &= 5\cdot 14\\ &= 70\mod 17\\ &= 2. \end{align}$$ The last step requires using the division algorithm to obtain $70 = 4(17)+2$ to get the remainder 2.
Performing the operations in the quotient ring $\mathbb{Z}/17\mathbb{Z}$ gives$$ \begin{align} 5(14+17\mathbb{Z}) &=5(-3+17\mathbb{Z})\\ &=-15+17\mathbb{Z}\\ &=2+ 17\mathbb{Z}, \end{align}$$ or the shortened version (dropping the implied $+17\mathbb{Z}$ from all terms)

$$5(14)=5(-3)=-15 = 2.$$ The last line above is lazy notation, but achieves the end result extremely fast. To make it not as lazy, you'll see things like $\equiv_{17}$ to supply the proper context. Unfortunately, humans get very lazy, which can lead to confusion and incorrect notation, because we start trying to use $\equiv_{17}$ for both equality in $\mathbb{Z}_{17}$ and equality in $\mathbb{Z}/17\mathbb{Z}$.

A More Complicated example.

Consider the groups $G=D_4\oplus U(3)$ and $H = \mathbb{Z}_2\times\mathbb{Z}_2$, along with the function $$ f(x,y) = \begin{cases} (0,y-1)& x\in\{R_0,R_{90},R_{180},R_{270}\} \\ (1,y-1)& x\notin\{R_0,R_{90},R_{180},R_{270}\} \end{cases} .$$ Essentially, the function $f$ maps $(x,y)$ to $(0,y-1)$ if $x$ is a rotation, and $(1,y-1)$ if $x$ is a flip. We can look at this a function on each coordinate, namely $f_1:D_4\to \mathbb{Z}_2$ and $f_2:U(3)\to\mathbb{Z}_2$, where $f_1(x)$ returns 0 if $x$ is a rotation or 1 if $x$ is a flip, and $f_2(y) = y-1$.

We will prove that $f$ is a homomorphism by first showing that both $f_1$ and $f_2$ are homomorphisms. Let $y_1,y_2\in U(3)$. Recall that $U(3) = \{1,2\}$ under multiplication mod 3, and $\mathbb{Z}_2 = \{0,1\}$ under addition mod 2. We first show that $f_2$ is a homomorphism, by showing that $ [ (y_1y_2\bmod 3)-1 ] \bmod 2 = [ (y_1-1)+(y_2-1) ]\bmod 2$ regardless of $x$. A brute force check shows that $$\begin{align} (1)(1)-1 &=0=(1-1)+(1-1),\\ (1)(2)-1 &= 1 = 0+1 = (1-1)+(2-1),\\ (2)(1)-1 &= 1 = 1+0 = (2-1)+(1-1),\\ (2)(2)\bmod{3}-1 &= 0 = 1+1 \bmod{2} = (2-1)+(2-1) \bmod{2}. \end{align}$$ This shows that $f_2(y_1y_2) = f_2(y_1)+f_2(y_2)$, so we know $f_2$ is a homomorphism.

Let's now focus on the first coordinate. Let $x_1,x_2\in D_4$. We must prove that $f_1(x_1\circ x_2) = f_1(x_1)+f_1(x_2)$. We will check every case.

If both $x_1$ and $x_2$ are rotations, then their composition is rotation which means $f_1(x_1\circ x_2) = 0 = 0+0= f_1(x_1)+f_1(x_2)$.
If both $x_1$ and $x_2$ are flips, then their composition is a rotation which means $f_1(x_1\circ x_2) = 0 = 1+1\bmod 2= f_1(x_1)+f_1(x_2)$.
If $x_1$ is a rotation and $x_2$ a flip, then their composition is flip which means $f_1(x_1\circ x_2) = 1 = 0+1= f_1(x_1)+f_1(x_2)$.
If $x_1$ is a flip and $x_2$ a rotation, then their composition is flip which means $f_1(x_1\circ x_2) = 1 = 1+0= f_1(x_1)+f_1(x_2)$.

In every case, we see that $f_1(x_1\circ x_2) = f_1(x_1)+f_1(x_2)$, which means $f_1$ is a homomorphism.

We now prove that $f(x,y) = (f_1(x),f_2(y))$ is a homomorphism. Let $(x_1,y_1),(x_2,y_2)\in G$, which means that $x_1,x_2\in D_4$ and $y_1,y_2\in U(3)$. We then compute $$\begin{align} f((x_1,y_1)\oplus_G(x_2,y_2)) &= f(x_1\circ x_2,y_1y_2) \\ &= (f_1(x_1\circ x_2),f_2(y_1 y_2)) \\ &= (f_1(x_1)+f(x_2),f_2(y_1)+f(y_2))\\ &= (f_1(x_1),f_2(y_1)) \oplus_H (f(x_2),f(y_2))\\ &= f(x_1,y_1)\oplus_H f(x_2,y_2). \end{align}$$

The kernel $K$ of $f$ is the set $(x,y)$ such that $x$ is a rotation and $y=1$. This set contains 4 elements. Note that $G$ has order 16, which means there are precisely 4 cosets of the kernel. These cosets are

$K = \{(R_0,1),(R_{90},1),(R_{180},1),(R_{270},1)\} $
$K(R_0,2) = \{(R_0,2),(R_{90},2),(R_{180},2),(R_{270},2)\} $
$K(V,1) = \{(V,1),(H,1),(D,1),(D',1)\} $
$K(V,2) = \{(V,2),(H,2),(D,2),(D',2)\} $

These 4 sets above form the factor group $G/K$. Note that the map $f$ is surjective, which means that that $G/K$ and $\mathbb{Z}_2\times\mathbb{Z}_2$ are isomorphic as groups.

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HomePage Current Problems All Problems Pictures ProblemsByWeek List of Problems List of Definitions List of Theorems List of Conjectures All Recent changes (use this to find a page that may have been lost)	Example Examples Of The First Isomorphism Theorem Please Login to access more options. The first isomorphism theorem states if $f:G\to H$ is a surjective homomorphism with kernel $K$, then the map $\phi:G/K\to H$ defined by $\phi(Kg)=f(g)$ is an isomorphism. We've been using this since the first weeks of the semester. Let's examine what this means in the context modular arithmetic. The group isomorphism $\mathbb{Z}/17\mathbb{Z}\approx \mathbb{Z}_{17}$ Let $G=\mathbb{Z}$ and $H = \mathbb{Z}_{17}$, and use the map $f: G\to H$ defined by $f(x) = x\mod 17$. Note that the kernel of $f$ is precise the multiples of 17, so we have $K=17\mathbb{Z}$. The first isomorphism theorem then says that $\mathbb{Z}/17\mathbb{Z}\approx \mathbb{Z}_{17}$. What is the difference? In $\mathbb{Z}_{17}$, the only elements are the integers from 0 to 16. No other integers exists, and all discussion about elements must refer to only these 17 elements. Each time we perform an operation, the mod operation is performed. In $\mathbb{Z}/17\mathbb{Z}$, we look at cosets. We know that $14+17\mathbb{Z} = -3+17\mathbb{Z}$ because $14-(-3)\in17\mathbb{Z}$. We use coset properties to perform computations and have no problem working with integers that are negative, or greater than 16. The mod operation is never performed, rather all elements that have the same remainder are just collected together in a single coset. No modding is required, rather we are free to pick the most convenient way to express the element for future computations. Let's look at these differences in terms of a computation. We will compute $14+(14+(14+(14+14)))) = 5(14)$. Performing the operations exclusively in $\mathbb{Z}_{17}$, we obtain $$ \begin{align} 14+(14+(14+(14+14)))) &= 14+(14+(14+11))) &&(28\mod 17 = 11)\\ &= 14+(14+8)) &&(25\mod 17 = 18)\\ &= 14+5 &&(22\mod 17 = 5)\\ &= 2 &&(19\mod 17 = 2). \end{align}$$ At each step above, we had to compute a sum, and then simplify mod 17. Performing the operations in $\mathbb{Z}/17\mathbb{Z}$ allows us to freely choose a convenient element in the coset. Since we know $14 + 17\mathbb{Z} = -3+17\mathbb{Z}$, we obtain $$ \begin{align} 5(14+17\mathbb{Z})&=14+ 17\mathbb{Z}+(14+ 17\mathbb{Z}+(14+ 17\mathbb{Z}+(14+ 17\mathbb{Z}+14+ 17\mathbb{Z}))))\\ &=-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+14+ 17\mathbb{Z}))))\\ &=-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(11+ 17\mathbb{Z})))\\ &=-3+ 17\mathbb{Z}+(-3+ 17\mathbb{Z}+(8+ 17\mathbb{Z}))\\ &=-3+ 17\mathbb{Z}+(5+ 17\mathbb{Z})\\ &=2+ 17\mathbb{Z}. \end{align}$$ The $+17\mathbb{Z}$ that appears on everything above is tedious to write, so often we just simply write $$5(14)=-3-3-3-3+14 = -3-3-3+11 =-3-3+8 =-3+5 = 2,$$ and then note that this gives the set equality $5(14+17\mathbb{Z}) = 2+17\mathbb{Z}$. The ring isomorphism $\mathbb{Z}/17\mathbb{Z}\approx \mathbb{Z}_{17}$ The computations above use the single binary operation + for both groups. When we add the binary operation $\cdot$ to the mix, we get ring structures. The first isomorphism theorem also holds for rings. This allows us to compute the product above much faster. We first note that $14+(14+(14+(14+14)))) = 5(14)$, so we can now use both multiplication and addition. Performing the operations exclusively in the ring $\mathbb{Z}_{17}$, we obtain $$ \begin{align} 14+(14+(14+(14+14)))) &= 5\cdot 14\\ &= 70\mod 17\\ &= 2. \end{align}$$ The last step requires using the division algorithm to obtain $70 = 4(17)+2$ to get the remainder 2. Performing the operations in the quotient ring $\mathbb{Z}/17\mathbb{Z}$ gives$$ \begin{align} 5(14+17\mathbb{Z}) &=5(-3+17\mathbb{Z})\\ &=-15+17\mathbb{Z}\\ &=2+ 17\mathbb{Z}, \end{align}$$ or the shortened version (dropping the implied $+17\mathbb{Z}$ from all terms) $$5(14)=5(-3)=-15 = 2.$$ The last line above is lazy notation, but achieves the end result extremely fast. To make it not as lazy, you'll see things like $\equiv_{17}$ to supply the proper context. Unfortunately, humans get very lazy, which can lead to confusion and incorrect notation, because we start trying to use $\equiv_{17}$ for both equality in $\mathbb{Z}_{17}$ and equality in $\mathbb{Z}/17\mathbb{Z}$. A More Complicated example. Consider the groups $G=D_4\oplus U(3)$ and $H = \mathbb{Z}_2\times\mathbb{Z}_2$, along with the function $$ f(x,y) = \begin{cases} (0,y-1)& x\in\{R_0,R_{90},R_{180},R_{270}\} \\ (1,y-1)& x\notin\{R_0,R_{90},R_{180},R_{270}\} \end{cases} .$$ Essentially, the function $f$ maps $(x,y)$ to $(0,y-1)$ if $x$ is a rotation, and $(1,y-1)$ if $x$ is a flip. We can look at this a function on each coordinate, namely $f_1:D_4\to \mathbb{Z}_2$ and $f_2:U(3)\to\mathbb{Z}_2$, where $f_1(x)$ returns 0 if $x$ is a rotation or 1 if $x$ is a flip, and $f_2(y) = y-1$. We will prove that $f$ is a homomorphism by first showing that both $f_1$ and $f_2$ are homomorphisms. Let $y_1,y_2\in U(3)$. Recall that $U(3) = \{1,2\}$ under multiplication mod 3, and $\mathbb{Z}_2 = \{0,1\}$ under addition mod 2. We first show that $f_2$ is a homomorphism, by showing that $ [ (y_1y_2\bmod 3)-1 ] \bmod 2 = [ (y_1-1)+(y_2-1) ]\bmod 2$ regardless of $x$. A brute force check shows that $$\begin{align} (1)(1)-1 &=0=(1-1)+(1-1),\\ (1)(2)-1 &= 1 = 0+1 = (1-1)+(2-1),\\ (2)(1)-1 &= 1 = 1+0 = (2-1)+(1-1),\\ (2)(2)\bmod{3}-1 &= 0 = 1+1 \bmod{2} = (2-1)+(2-1) \bmod{2}. \end{align}$$ This shows that $f_2(y_1y_2) = f_2(y_1)+f_2(y_2)$, so we know $f_2$ is a homomorphism. Let's now focus on the first coordinate. Let $x_1,x_2\in D_4$. We must prove that $f_1(x_1\circ x_2) = f_1(x_1)+f_1(x_2)$. We will check every case. If both $x_1$ and $x_2$ are rotations, then their composition is rotation which means $f_1(x_1\circ x_2) = 0 = 0+0= f_1(x_1)+f_1(x_2)$. If both $x_1$ and $x_2$ are flips, then their composition is a rotation which means $f_1(x_1\circ x_2) = 0 = 1+1\bmod 2= f_1(x_1)+f_1(x_2)$. If $x_1$ is a rotation and $x_2$ a flip, then their composition is flip which means $f_1(x_1\circ x_2) = 1 = 0+1= f_1(x_1)+f_1(x_2)$. If $x_1$ is a flip and $x_2$ a rotation, then their composition is flip which means $f_1(x_1\circ x_2) = 1 = 1+0= f_1(x_1)+f_1(x_2)$. In every case, we see that $f_1(x_1\circ x_2) = f_1(x_1)+f_1(x_2)$, which means $f_1$ is a homomorphism. We now prove that $f(x,y) = (f_1(x),f_2(y))$ is a homomorphism. Let $(x_1,y_1),(x_2,y_2)\in G$, which means that $x_1,x_2\in D_4$ and $y_1,y_2\in U(3)$. We then compute $$\begin{align} f((x_1,y_1)\oplus_G(x_2,y_2)) &= f(x_1\circ x_2,y_1y_2) \\ &= (f_1(x_1\circ x_2),f_2(y_1 y_2)) \\ &= (f_1(x_1)+f(x_2),f_2(y_1)+f(y_2))\\ &= (f_1(x_1),f_2(y_1)) \oplus_H (f(x_2),f(y_2))\\ &= f(x_1,y_1)\oplus_H f(x_2,y_2). \end{align}$$ The kernel $K$ of $f$ is the set $(x,y)$ such that $x$ is a rotation and $y=1$. This set contains 4 elements. Note that $G$ has order 16, which means there are precisely 4 cosets of the kernel. These cosets are $K = \{(R_0,1),(R_{90},1),(R_{180},1),(R_{270},1)\} $ $K(R_0,2) = \{(R_0,2),(R_{90},2),(R_{180},2),(R_{270},2)\} $ $K(V,1) = \{(V,1),(H,1),(D,1),(D',1)\} $ $K(V,2) = \{(V,2),(H,2),(D,2),(D',2)\} $ These 4 sets above form the factor group $G/K$. Note that the map $f$ is surjective, which means that that $G/K$ and $\mathbb{Z}_2\times\mathbb{Z}_2$ are isomorphic as groups. Edit Page History Source Attach File Backlinks List Group Page last modified on December 13, 2019, at 01:50 PM
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Examples Of The First Isomorphism Theorem

The group isomorphism $\mathbb{Z}/17\mathbb{Z}\approx \mathbb{Z}_{17}$

The ring isomorphism $\mathbb{Z}/17\mathbb{Z}\approx \mathbb{Z}_{17}$

A More Complicated example.