If $n$ is an odd and positive integer, show that $D_{4n}\cong D_{2n}\,\times\, Z_2$. Denote that $D_{2n}$ is the Dihedral Groups with $2n$ elements and $Z_2$ is a cyclic group with order 2. I think it might be solved by using the recognizing theorem.
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4This is going to be difficult, because it's false. For instance, $D_4$ is not isomorphic to $D_2\times \mathbb{Z}_2$. – Captain Lama Apr 26 '20 at 17:17
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Then what if $n\geq3$? – 廖倪徵 Apr 26 '20 at 17:45
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Try to look at the maximal order of an element in $D_{4n}$. – Captain Lama Apr 26 '20 at 17:57
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This is false efor every $n$. Since every dihedral $D_{2n}$ has a cyclic subgroup of maximal order $2n$, $D_{2n} \times Z_2$ has the its biggest cyclic group of the same order, while $D_{4n}$ has a cyclic subgroup of order $4n$.
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If we can find $D_{4n}=HK$ where $H, K$ are normal subgroups of $D_{4n}$ satisfying $H\cong D_{2n}$ and $K\cong Z_2$, $H\cap K=e$? – 廖倪徵 Apr 26 '20 at 18:13
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@廖倪徵 You won’t be able to; you can find such groups, but not both normal. $H$ will be normal (index $2$), but you won’t find a normal complement. It just isn’t true except for the dihedral group of order $4$. – Arturo Magidin Apr 26 '20 at 20:06