Are these two quotient groups of $U(16)$ isomorphic?

Let $G=U(16)$, $H=\{1,15\}$, $K=\{1,9\}$. Are $H$ and $K$ isomorphic? What about $G/H$ & $G/K$?

I found that $H$ and $G$ are isomorphic as both are cyclic and every finite cyclic group is isomorphic to group $Z_n$.

Here both $H$ and $G$ are isomorphic to $Z_2$. So both $H$ and $G$ are isomorphic. Now as $H$ and $G$ are cyclic, can I conclude that the quotient groups are abelian?

• Any quotient of an abelian group is also abelian. But that doesn't mean that $G/H$ and $G/K$ are isomorphic. (I don't know whether they are or not.) – Matt Samuel Dec 2 '16 at 7:05
• Here it is asked what can we conclude about the factor groups? H and K are surely isomorphic. – Kavita Dec 2 '16 at 7:15
• I agree that $H$ and $K$ are isomorphic. But that doesn't automatically imply that $G/H$ and $G/K$ are isomorphic. – Matt Samuel Dec 2 '16 at 7:19

1 Answer

(Note that almost every time you mention $H$ and $G$, you really mean $H$ and $K$. I will continue to refer to the two subgroups as $H$ and $K$).

As $U(16)$ is abelian, any subgroup is abelian. This includes quotient groups.

The group $U(16) = \{1,3,5,7,9,11,13,15\}$ and has eight elements. As $H$ and $K$ each have $2$ elements, we know that $G/H$ and $G/K$ will each have $4$ elements. Let's find the quotient groups explicitly, as it's a good exercise in understanding quotient groups.

First, let's find $G/H$. $$\begin{array}{lll} 1\cdot \{1,15\} &= \{1,15\} &= 15 \cdot \{1,15\} \\ 3\cdot \{1,15\} &= \{3,13\} &= 13 \cdot \{1,15\} \\ 5 \cdot\{1,15\} &= \{5,11\} &= 11 \cdot \{1,15\} \\ 7 \cdot \{1,15\} &= \{7,9\} &= 9 \cdot \{1,15\} \end{array}$$ This group is cyclic of order $4$, and the coset $3 \cdot \{1,15\}$ is a generator.

Now let's find $G/K$. $$\begin{array}{lll} 1\cdot\{1,9\} &= \{1,9\} &= 9\cdot\{1,9\} \\ 3\cdot\{1,9\} &= \{3,11\} &= 11\cdot\{1,9\} \\ 5\cdot\{1,9\} &= \{5,13\} &= 13\cdot\{1,9\} \\ 7 \cdot \{1,9\} &= \{7, 15\} &= 15\cdot\{1,9\} \end{array}$$ Notice that $(3 \cdot\{1, 9\})^2 = 9 \cdot \{1, 9\} = \{1,9\}$, and $(5 \cdot\{1,9\})^2 = 9 \cdot \{1,9\} = \{1,9\}$, and $(7\cdot\{1,9\})^2 = 1 \cdot \{1,9\} = \{1,9\}$. So the square of every coset is the identity.

Thus $G/K$ is not cyclic, and so the two quotient groups are not isomorphic. (In fact, note that in $G/H$, we have that $3 \{1,9\}$ times $5 \{1,9\}$ is $15\{1,9\} = 7\{1,9\}$, so in fact $G/H$ is isomorphic to the Klein four group, not the cyclic group). $\spadesuit$