$D_3\oplus D_4$ not isomorphic to $D_{24}$ We need to prove that $D_{3} \oplus D_{4}$ is not isomorphic to $D_{24}$ .
The way in which I approach such type of questions is to count the number of elements of order $x$ in one group and then in the other group, and then conclude that they aren't equal and hence there can't be any isomorphism between them.
But this approach here doesn't look easy, with $D_{24}$ involved.
Could anyone suggest another way of solving this ?
 A: Hint The group $D_{24}$ (usually denoted $D_{48}$) has an element of order $24$. Does $D_3 \times D_4$?
A: Here is a stronger result (related to this thread).  I claim that the dihedral group $D_n$ is decomposable if and only if $n\equiv 2\pmod{4}$.  (In particular, when $n=24$, we have $n\not\equiv 2\pmod{4}$, so $D_n=D_{24}$ is indecomposable, so you can't write $D_{24}=D_3\times D_4$.)  The case $n\in\{1,2\}$ is easy.  We now assume that $n\geq 3$.  
If $D_n=A\times B$ for some non-trivial subgroups $A$ and $B$, then $A$ and $B$ are normal subgroups of $D_n$, and $A\cap B=\{1\}$.  Write $D_n=\langle R,F\rangle$, where $R^n=F^2=1$ and $(RF)^2=1$ as given here.  Then, the normal subgroups of $D_n$ are


*

*$n$ odd: $\langle R^d\rangle$ where $d$ is a divisor of $n$;

*$n$ even: $\langle R^d\rangle$ where $d$ is a divisor of $n$, and two more $\langle R^2,F\rangle$ and $\langle R^2,RF\rangle$.
So, if $n$ is odd, then both $A$ and $B$ are subgroups of the cyclic subgroup $\langle R\rangle$, which means $A$ and $B$ are abelian, and so $D_n=A\times B$ is abelian, which is a contradiciton.
If $n\geq 4$ is even, then exactly one of $A$ and $B$ takes the form $\langle R^2,F\rangle$ or $\langle R^2,RF\rangle$.  WLOG, $A$ is of such a form.  Then, $A\cong D_{n/2}$.  So, $B$ must be isomorphic to the cyclic group $C_2$ of order $2$.  
If $n\equiv 2\pmod{4}$, then we can take $A=\langle R^2,F\rangle$ and $B=\langle R^{n/2}\rangle$ to see that $$D_n=A\times B\cong D_{n/2}\times C_2.$$  In fact, there are exactly two ways to decompose $D_n$ into a direct product of non-trivial subgroups:
$$D_n=\langle R^2,F\rangle \times \langle R^{n/2}\rangle$$
and
$$D_n=\langle R^2,RF\rangle \times \langle R^{n/2}\rangle.$$
If $4\mid n$, 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 $4\mid n$.
A: HINT: Look at the commutator subgroups of elements of order three in both groups.
