Question

Show that ${\mathit{U}}_{\mathbf{32}}\mathbf{/}\mathit{N}\mathbf{\cong }{\mathit{U}}_{\mathbf{16}}$, where N is the subgroup $\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{17}\mathbf{\}}$.

Short answer

It is proved that $U_{32}/N$is isomorphic to $U_{16}$.

Step-by-step solution

Proving homomorphism between U32/N and U16

Let’s consider, $\varphi :U_{32}/N\to U_{16}$

So, it can be defined as,

$\varphi \left(a_{32}\right)=\left[a_{16}\right]$

For any arbitrary a, b:

$$\begin{gathered} \varphi \left(\left(a_{16}\right)\left(b_{32}\right)\right)=\varphi \left(\left(a_{32}{b}_{32}\right)\right) \\ =\left[a_{16}{b}_{16}\right] \\ =\left[a_{16}\right]\left[b_{16}\right] \\ =\varphi \left(a_{16}\right)\varphi \left(b_{16}\right) \end{gathered}$$

Therefore, it is a homomorphism.

One to one mapping

Since$U_{16}\subset U_{32}$

As we know 17 is not an element in $U_{16}$, so it is mapped to 1 by a subgroup of $N=\{1,17\}$.

So, one-to-one mapping exists between both groups.

Order of groups

Since there are 8 elements in $U_{16}$the order $U_{16}$of is 8.

The order of $U_{32}/N$can be given as:

$\begin{array}{c}{U}_{32}/N=\left|U_{32}\right|/\left|N\right|\\ =16/2\\ =8\end{array}$

Therefore, both the groups have the same order.

Since both the groups satisfy all the conditions for isomorphism. Hence, it is proved that $U_{32}/N$is isomorphic to $U_{16}$.