Question

Prove that ${\mathbf{U}}_{\mathbf{16}}$is isomorphic to ${\mathbf{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_4$.[Hint: Theorem 9.3.]

Short answer

Answer:

It is proved that, $U_{16}$is isomorphic to ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_4$.

Step-by-step solution

Find the multiplicative group U16

Definition of subgroup:

Let $\left(G,*\right)$be a group under binary operation, say *. A non-empty subset $H$of $G$is said to be a subgroup of $G$, if $\left(H,*\right)$is itself a group.

In other words, if $e_G\in H$and $a,b\in H$then $a{b}^{-1}\in H$.

Definition of Normal Subgroup:

A subgroup $N$of a group $G$is called a normal subgroup of $G$if $Na=aN,\forall a\in G$.

Theorem 9.3can be stated as follows:

If $M$and $N$are normal subgroups of a group $G$such that $G=MN$and , then $G=M\times N$.

The multiplicative group of units in ${\mathrm{\mathbb{Z}}}_{16}$is $U_{16}$i s given by,$U_{16}=\left\{1,3,5,7,9,11,13,15\right\}$.

We have to prove that $U_{16}$is isomorphic to ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_4$.

Prove U16 is isomorphic toℤ2×ℤ4

Let $A=\left\{7^1=7,7^2=1\right\}$and $B=\left\{3^1=3,3^2=9,3^3=11,3^4=1\right\}$.

Then, $A=\left\{1,7\right\},B=\left\{1,3,9,11\right\}$, and $A\cap B=\left\{1\right\}$.

Also, $U_{16}$can be written as:

$$\begin{gathered} 9=1*9 \\ 11=1*11 \\ 13=7*11 \\ 15=7*9 \end{gathered}$$

Further, as:

$$\begin{gathered} 9=1*9 \\ 11=1*11 \\ 13=7*11 \\ 15=7*9 \end{gathered}$$

Hence, $U_{16}=AB$.

Since $U_{16}$is abelian, $A,B$are normal subgroups.

Thus, byTheorem 9.3,$U_{16}=A\times B={\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_4.$

Hence, $U_{16}$is isomorphic to ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_4$.