On the centres of the dihedral groups
In an evidence that I lately read, the adhering to 'reality' is made use of, where $D_{2n}$ represents the dihedral team of order $2n$:
If $n$ is also, after that $D_{2n} \cong C_2 \times D_n$.
The (brief) offered validation is that the centre $Z(D_{2n}) \cong C_2$, whenever $n$ is also, and also is unimportant given that $n$ is weird.
Nonetheless, below is an outcome that a close friend of mine located in the litterature which negates the previous one.
Intend that 4 separates $n$. After that $D_{2n}$ is not isomorphic to $C_2 \times D_n$.
Evidence: Suppose or else. We understand that $Z(D_{2n}) \cong C_2$. By presumption, $D_{2n} \cong C_2 \times D_n$, consequently $Z(D_{2n}) \cong Z(C_2 \times D_{n})$. Given that in a straight item $A \times B$, the teams $A$ and also $B$ commute, we get that $Z(C_2 \times D_{n}) = Z(C_2) \times Z(D_n) \cong C_2 \times C_2$, an opposition, QED
Now comes my first inquiry : Is the above evidence deal with?
2nd inquiry : Given a limited team $G$ whose centre is not unimportant, does it constantly exist a team $H$ such that $G \cong Z(G) \times H$?
If so, after that offered $n$ a multiple of 4, the dihedral team $D_{2n}$ can be created as a straight item $C_2 \times H$. Below comes my 3rd inquiry : What should be $H$ given that we understand that it can not have nontrivial centre (and also, specifically, given that $H$ can not be $D_n$)?
I like your inquiries.

Yes, and also it is really foolish.

No, as an example the quaternion team of order 8 is not like this.

(So, there is no such H).
To be clear, the dihedral team of order 8k+4 is isomorphic to the straight item of a cyclic team of order 2 with a dihedral team of order 4k+2, yet all various other dihedral groups are "directly indecomposable", that is, they do not have a disintegration as a straight item of 2 non  identification teams.
In instance you intend to confirm the more powerful indecomposability: Basically, you consider the regular subgroups. Thinking the order of the dihedral team goes to the very least 6, they are all had in the subgroup of turnings, therefore the only means you can have a straight item is if that subgroup of turnings has order 4k+2, to make sure that its Sylow 2  subgroup is systematized by a flip.
Question 1. The evidence is a proper evidence that the initial assertion (that for $n$ also you have $D_{2n}\cong C_2\times D_n$) is incorrect. Without a doubt, you can merely do it with $n=4$: $D_4$ is abelian, the Klein $4$  team ; $D_8$ is not abelian, yet $C_2 \times D_4$ is abelian, so $D_8$ can not be isomorphic to $C_2\times D_4$. The assertion concerning $D_{2n}$ is incorrect.
Inquiry 2. No, it is not real as a whole. Take a nonabelian team of order $p^3$ ; after that $Z(G)\cong C_p$. If you had $G\cong Z(G)\times H$, after that $H$ would certainly be of order $p^2$, therefore abelian, so $G$ would certainly be abelian (an item of 2 abelian groups).
Inquiry 3. Moot.
A comment concerning the assertion (first inquiry).
"when n is also after that $D_{2n}\cong C_2 \times D_{n}$" .
This seems not real : take into consideration an unique class of dihedral groups particularly $2$groups, i.e. dihedral groups of order $2^n$, with $n\geq 3$. These are non  abelian $2$ teams. As you claimed, their facilities are non  unimportant (need to be ; it holds true for any kind of p  team) ; the facility of the dihedral team $D_{2^n}$ of order $2^n$ is $C_2$.
Yet by Sylow  concept, it needs to converge with every regular subgroup of $D_{2^n}$ non  trivially, therefore need to be had because regular subgroup (as facility has order 2) ; specifically the facility have to be had in every topmost subgroup of $D_{2^n}$ , and also therefore specifically, it needs to be had in subgroups isomorphic to $D_{2^{n1}}$.
Consequently, $D_{2^n}$ can not be (inner) semidirect item of $C_2$ by $D_{2^{n1}}$. ; therefore specifically it can be straight item of $C_2$ by $D_{2^{n1}}$.
[An additional straightforward argument can be offered: if for $n$ also, $D_{2n}\cong C_2 \times D_{n}$, after that specifically we need to have (by induction)
$D_{2^n} \cong C_2 \times D_{2^{n1}} \cong C_2 \times (C_2 \times D_{2^{n2}})\cong \cdots \cong C_2 \times C_2 \times \cdots \times D_4$, which is primary abelian $2$  team, opposition.
Currently really straightforward solutions offered over to 2nd and also 3rd inquiry, so absolutely nothing to add even more. ]
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