0
$\begingroup$

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.

Thanks in advance.

$\endgroup$
  • 2
    $\begingroup$ I think you need to calculate the actual groups! $\endgroup$ – Derek Holt Nov 15 '14 at 11:25
1
$\begingroup$

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 $>2$.

$\endgroup$
2
$\begingroup$

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,

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.