Isomorphism type of two quotient groups of $G=\mathbb{Z}^\times_{16}$

I'm supposed to determine the isomorphism type of $G/\langle 15\rangle$ and $G/\langle 9 \rangle$.

I've determined the order of both subgroups ($\langle 15\rangle$ and $\langle 9\rangle$), and it is $2$. Therefore, the order of quotient groups is $4$ and I'm left with two possibilities for the isomorphism type of each quotient group i.e. $Z_4$ and $Z_2 \times Z_2$. However, I'm a bit stuck on how to determine which is which other than simply calculating the actual groups (Which I would like to avoid). I've realized that all the groups in the problem are solvable ($p$-groups) so I guess I should use this somehow, but I'm not really sure how!

Any help would be well appreciated.

• I think you need to calculate the actual groups! – Derek Holt Nov 15 '14 at 11:25

Make yourself clear that $G/\langle 9\rangle\cong \mathbb Z_8^\times$. Or: Since the square of an odd integer is always $\equiv 1\pmod 8$, it is $\equiv 1\pmod{16}$ or $\equiv 9\pmod {16}$. Hence all elements of $G/\langle 9\rangle$ have order $\le2$.

On the other hand, $3^2\notin\langle15\rangle$, hence $G/\langle 15\rangle$ has an element of order $>2$.

You just have to compute the squares of the elements (three of them, so it is not a lot of computations). The group $G$ has order $8$ and is abelian, so it is either $Z_2 \times Z_2 \times Z_2$, $Z_4 \times Z_2$ or $Z_8$. We can already eliminate $Z_2 \times Z_2 \times Z_2$ since $3^2 = 9 \neq 1$. Since you can check that $x^4 = 1$ for all $x \in \mathbb Z_{16}^{\times}$ (just square $3,5,7$, since $9^2 = (-7)^2 = 7^2$, and so on - then you will see), you see that $Z_8$ is also impossible, thus you have $Z_4 \times Z_2$.

Since our elements $15$ and $9$ are of order two, they correspond to either $(2,0)$ or $(0,1)$ in $Z_4 \times Z_2$. The first quotient gives $Z_2 \times Z_2$ and the second quotient gives $Z_4$, so we need to be careful (as I wasn't the first time, mostly because I didn't pay enough attention).

The group $Z_4$ is cyclic, but $Z_2 \times Z_2$ is not. If you can find an element of order $4$ in the group $G$ whose square is $15$ (or $9$), then the corresponding group is cyclic of order $4$ ; otherwise it is isomorphic to $Z_2 \times Z_2$. I leave the (very minor) calculations up to you.

Hope that helps,

• But $G$ doesn't have any elements of order $8$. – Derek Holt Nov 15 '14 at 11:30
• @DerekHolt : Then... =) – Patrick Da Silva Nov 15 '14 at 12:13
• But I think you will find that the $H/\langle 15 \rangle$ is $Z_4$. – Derek Holt Nov 15 '14 at 21:34
• @DerekHolt : Bleh, I don't pay enough attention on this website anymore... I guess I forgot about $Z_4 \times Z_2$. – Patrick Da Silva Nov 15 '14 at 22:59
• @Derek Holt : Now this is better. – Patrick Da Silva Nov 15 '14 at 23:10