Htykuut

How to show $D_3oplus D_4$ is not isomorphic to $D_{24}$?

Here $D_n$ is the dihedral group of order $2n$.

I am not sure how to prove this. I am not very good with the dihedreal groups.

Hint The group $D_{24}$ (usually denoted $D_{48}$) has an element of order $24$. Does $D_3 times D_4$?

Here is a stronger result (related to this thread). I claim that the dihedral group $D_n$ is decomposable if and only if $nequiv 2pmod{4}$. (In particular, when $n=24$, we have $nnotequiv 2pmod{4}$, so $D_n=D_{24}$ is indecomposable, so you can't write $D_{24}=D_3times D_4$.) The case $nin{1,2}$ is easy. We now assume that $ngeq 3$.

If $D_n=Atimes B$ for some non-trivial subgroups $A$ and $B$, then $A$ and $B$ are normal subgroups of $D_n$, and $Acap B={1}$. Write $D_n=langle R,Frangle$, where $R^n=F^2=1$ and $(RF)^2=1$ as given here. Then, the normal subgroups of $D_n$ are

• $n$ odd: $langle R^drangle$ where $d$ is a divisor of $n$;




• $n$ even: $langle R^drangle$ where $d$ is a divisor of $n$, and two more $langle R^2,Frangle$ and $langle R^2,RFrangle$.

So, if $n$ is odd, then both $A$ and $B$ are subgroups of the cyclic subgroup $langle Rrangle$, which means $A$ and $B$ are abelian, and so $D_n=Atimes B$ is abelian, which is a contradiciton.

If $ngeq 4$ is even, then exactly one of $A$ and $B$ takes the form $langle R^2,Frangle$ or $langle R^2,RFrangle$. WLOG, $A$ is of such a form. Then, $Acong D_{n/2}$. So, $B$ must be isomorphic to the cyclic group $C_2$ of order $2$.

If $nequiv 2pmod{4}$, then we can take $A=langle R^2,Frangle$ and $B=langle R^{n/2}rangle$ to see that $$D_n=Atimes Bcong D_{n/2}times C_2.$$ In fact, there are exactly two ways to decompose $D_n$ into a direct product of non-trivial subgroups:
$$D_n=langle R^2,Frangle times langle R^{n/2}rangle$$
and
$$D_n=langle R^2,RFrangle times langle R^{n/2}rangle.$$

If $4mid n$, then all elements of order $2$ of $D_n$ are $R^{n/2}$ and $R^kF$. But the subgroup generated by $R^kF$ is not normal, and $R^{n/2} in A$ because $frac{n}{2}$ is even. Therefore, $A$ and $B$ cannot exist. Hence, $D_n$ is indecomposable when $4mid n$.

HINT: Look at the commutator subgroups of elements of order three in both groups.