In my notation $$D_{12}=\langle \rho,\tau : \rho^6=\tau^2=1,\ \rho \tau \rho=\tau\rangle$$ So firstly, I know that all characteristic subgroups are normal. Thus, the possible candidates of $D_{2n}$ for the characteristic subgroups are $$\{1\},D_{12},<\rho>,<\rho^2>,<\rho^3>,<\rho^2,\tau>,<\rho^2,\rho\tau>$$ Is there now a property or tool which I can use to pick out the characteristic subgroups. I know for sure that $D_{12}$ and $\{1\}$ as well as the commutator subgroup $D_{12}'=<\rho>$ are characteristics, but what about the others.


For starters, consider the orders of the subgroups. If there is only one subgroup of a certain order, surely it must be characteristic. Of the three remaining subgroups, only one is cyclic, and so it must also be characteristic. For the remaining two groups, it is not hard to check whether there is an automorphism mapping one to the other.

  • $\begingroup$ Can you help me to list the elements of the subgroup of $<\rho^2,\rho \tau>$. It's all possibgle products of $\rho ^2$ and $\rho \tau$, right? $\endgroup$ – zermelovac Aug 21 '18 at 19:50
  • $\begingroup$ Yes, it is. And I'm sure you can list them yourself, there are only six elements. $\endgroup$ – Servaes Aug 21 '18 at 19:53
  • $\begingroup$ I gor $<\rho^2,\rho \tau >=\{1,\rho^2,\rho^4,\rho \tau, \rho^3\tau, \rho^5\tau\}$ $\endgroup$ – zermelovac Aug 21 '18 at 19:57
  • $\begingroup$ So, since $<\rho^2>,<\rho^3> and <\rho^2,\tau>$ have respectively orders 3,2 and 5, they are characterstics. I just need to check for the last one? $\endgroup$ – zermelovac Aug 21 '18 at 19:59
  • $\begingroup$ The first two groups; yes. The group $\langle\rho^2,\tau\rangle$ does not have order 5. In general, the order of a subgroup divides the order of the group. $\endgroup$ – Servaes Aug 21 '18 at 20:00

As for the remaining $2$ subgroups, $\langle \rho^2,\tau\rangle$ and $\langle \rho^2,\rho\tau\rangle$ there is an automorphism of $D_{12}$ sending them to each other... Namely, the one determined by $\rho\to\rho$ and $\tau\to\rho\tau$. (By the relation $\rho\tau\rho=\tau$, the elements $\tau$ and $\rho\tau$ both have order $2$; or, they are both reflections.)

So, these two subgroups aren't characteristic...


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