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If $G \cong H \times \mathbb{Z}_2, $ show that $G$ contains an element $a$ of order $2$ with the property that $ag = ga$ for all $g \in G$. Deduce (briefly!) that the dihedral group $D_{2n+1}$ (with $n \geq 1$) is not isomorphic to a product $H \times C_2$.

Attempt:

The isomorphism implies a bijective homomorphism of the form $\phi: G \rightarrow H \times \mathbb{Z}_2$ with $g \mapsto (h,z),$ $g$ in $G$, $h$ in $H$ and $z$ in {$0,1$}. This means $|G| = 2|H|$ and so the order of $G$ must be even, $|G| = 2k,$ for $k$ in naturals.

The order of elements of $G$ (or the order of any subgroups) must divide the order of the group by Lagrange and hence all elements in $G$ have even order. Elements are of order $g^{2\alpha}, \alpha \in \mathbb{N}$ and each creates a cyclic subgroup. (Right?) So the possible orders of alpha are $2,4,6..2α$. So $G$ does have an element of order $2$. Since the subgroup generated by this element is prime, necessarily it is cyclic which in turns implies the group is abelian. This subgroup is a subset of $G$, so the commutavity is 'maintanied' through the group and so $ag=ga.$

Suppose there was an isomorphism between $D_{2n+1}$ and $H \times C_2$. This means $|H| = 2n+1$, $n$ nonnegative. So the order of $H$ is odd and the order of $C_2$is even and so the orders are coprime. This means $H \times C_2$ is cyclic. Can I deduce anything about $D_{2n+1}$ from this?

Many thanks.

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"the commutativity is maintained through the group"? What does this mean exactly...and why is it true? –  DonAntonio Apr 30 '13 at 21:55
    
Yes, I was a bit unsure about this statement. What I meant was that given that some subgroup is abelian then the whole group is abelian. I think this is true since a subgroup of a group will have all the properties of the original group. –  CAF Apr 30 '13 at 22:00
    
Well, that is blatantly false, of course, and in a very explosive way: all the proper subgroups of $\,S_3\,$ are abelian yet $\,S_3\,$ itself isn't... –  DonAntonio Apr 30 '13 at 22:03
    
I see, so how should I prove this? –  CAF Apr 30 '13 at 22:09
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2 Answers

Beyond what DonAntonio has already pointed out, there are a few more issues with your argument.

You're right that $G$ has even order, and that the order of every subgroup divides this order, but this does not imply the order of every subgroup is even. For example if the order of $G$ is 6, this is even, $3$ divides it, and so $G$ may have a subgroup of order $3$. If, for example, $G = C_3\times C_2$, then this happens. (In fact, it happens for the other group of order $6$ as well.)

Even if you knew all subgroups have even order, and thus, that all elements have even order, this doesn't guarantee an element of all even orders exists. (In the particular case of prime order, it turns out such an element will exist - this goes by the name Cauchy's theorem. So, your conclusion is right, even if the argument itself is not right.)

Now, with that out of the way, here's a two part hint for the first part.

Hint 1: Prove that for any group $H$, $H\times C_2$ contains an element of order $2$, called $a$, for which $ag=ga$ for all $g\in H$.

Hint 2: If $G$ and $H\times C_2$ are isomorphic, let $f:G\rightarrow H\times C_2$ be a particular isomorphism. Assuming hint $1$, we know $H\times C_2$ has a special element $a$. How can you use $f$ to turn this into a special element in $G$? Can you prove this special element in $G$ has the desired properties?

For the second part of your question, try to argue that no element of order $2$ in $D_{2n+1}$ commutes with everything. More generally, show that no element of $D_{2n+1}$, regardless of order, commutes with everything in $D_{2n+1}$.

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The claim is true if for example $\;2\nmid |H|\;$ , since in this case there is only one element of order $\,2\,$ (=an involution) which is then central (why? Look at the conjugates of the involution).

Added: If a group $\,G\,$ has one unique element $\,a\,$ of order two then the group $\,\langle a\rangle =\{1,a\}\,$ is central.

Proof: We have that

$$\forall\,x\in G\;,\;\;\text{ord}(a^x:=x^{-1}ax)=\text{ord}(a)\,\implies$$

$$\implies \begin{cases}a^x=1\iff a=1\\{}\\a^x=a\iff ax=xa\end{cases}\;\;\implies\langle a\rangle\le C(G)\;\;\;\;\;\;\;\;\;\;\square$$

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I haven't covered involutions and centralizers were right at the end of my course. This exercise came from sections about Isomorphisms and Products. Is there a way to use the material there? Thanks. –  CAF Apr 30 '13 at 22:20
    
@CAF , check the added stuff in my answer. –  DonAntonio Apr 30 '13 at 22:31
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