# Thread: A normal sylow 3-subgroup of D6

1. ## A normal sylow 3-subgroup of D6

I could not verify the below paragraph in my textbook [Hungerford, p98].

Let $c \in G$, G is a group of order 12, and P is the normal (unique) sylow 3-subgroup. Hence G contains only two elements of order 3. If c is one of these, then $[G:C_{G}(c)]=1$ or $2$. Thus $C_{G}(c)$ is a group of order 12 or 6. In either case there is $d \in C_{G}(c)$ of order 2 by Cauchy's theorem. Verify that |cd| = 6.

In my textbook, $D_{6}$ is such a group that has a normal sylow 3-subgroup.

$D_{6} = \{1, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5\}$.
What is the normal sylow 3-subgroup of the above $D_{6}$?

2. Originally Posted by aliceinwonderland
I could not verify the below paragraph in my textbook [Hungerford, p98].

Let $c \in G$, G is a group of order 12, and P is the normal (unique) sylow 3-subgroup. Hence G contains only two elements of order 3. If c is one of these, then $[G:C_{G}(c)]=1$ or $2$. Thus $C_{G}(c)$ is a group of order 12 or 6. In either case there is $d \in C_{G}(c)$ of order 2 by Cauchy's theorem. Verify that |cd| = 6.
If $a,b\in G$ are elements that commute (with finite orders) then $|ab| = \text{lcm}(|a|,|b|)$.
Now $|c|=3$ and $|d|=2$. However, $cd=dc$ because by definition $d\in C_G(c)$.
The rest follows.

3. Originally Posted by ThePerfectHacker
If $a,b\in G$ are elements that commute (with finite orders) then $|ab| = \text{lcm}(|a|,|b|)$.
unfortunately this is not always true, even if we assume the group is abelian. (for example, think about when $a=b$) we also need the orders to be coprime, which in this case: $|ab|=|a||b|.$

however, it is true that there always exists an element of G of order $\text{lcm}(|a|,|b|).$ this is a good problem which is worth thinking!

for the second part of aliceinwonderland's question: the Sylow 3-subgroup of $D_6$ is: $=\{1,r^2,r^4 \}.$

4. Originally Posted by NonCommAlg
unfortunately this is not always true, even if we assume the group is abelian. (for example, think about when $a=b$) we also need the orders to be coprime, which in this case: $|ab|=|a||b|.$
Yes , what was I thinking! But lucky my argument works because $|ab| = |a||b|$ as you said.

however, it is true that there always exists an element of G of order $\text{lcm}(|a|,|b|).$ this is a good problem which is worth thinking!
I posted a solution a year ago, here via Sylow.

5. Originally Posted by ThePerfectHacker

I posted a solution a year ago, here via Sylow.
the problem i gave you was slightly different. it was about cyclic subgroups. here is the complete question:

let G be a group (abelian or non-abelian) and $a,b \in G$ two elements of finite order which commute. show that G has an element of order $\text{lcm}(|a|,|b|).$

6. Originally Posted by aliceinwonderland
Let $c \in G$, G is a group of order 12, and P is the normal (unique) sylow 3-subgroup. Hence G contains only two elements of order 3. If c is one of these, then $[G:C_{G}(c)]=1$ or $2$. Thus $C_{G}(c)$ is a group of order 12 or 6. In either case there is $d \in C_{G}(c)$ of order 2 by Cauchy's theorem. Verify that |cd| = 6.
Thanks for all replies.

I have one last paragraph in my text that I could not understand regarding a nonabelian group of order 12.

"Let a = cd (in the above quote); then <a> is normal in G and |G/<a>| = 2. Hence there is an element $b \in G$ such that $b \notin , b \neq e, b^{2} \in , bab^{-1} \in $.
Since G is nonabelian and $|a|=6, bab^{-1} = a^{5} = a^{-1}$ is the only possibility.
....
There are six possibilities for $b^{2} \in . b^{2}=a^{2}, b^{2}=a^{4}$ lead to contradictions. $b^{2}=a$ or $b^{2}=a^{5}$ imply |b|=12.
..........."

Why the $b^{2}=a^{2}, b^{2}=a^{4}$ lead to contradiction?

7. Originally Posted by aliceinwonderland
Thanks for all replies.

I have one last paragraph in my text that I could not understand regarding a nonabelian group of order 12.

"Let a = cd (in the above quote); then <a> is normal in G and |G/<a>| = 2. Hence there is an element $b \in G$ such that $b \notin , b \neq e, b^{2} \in , bab^{-1} \in $.
Since G is nonabelian and $|a|=6, bab^{-1} = a^{5} = a^{-1}$ is the only possibility.
....
There are six possibilities for $b^{2} \in . b^{2}=a^{2}, b^{2}=a^{4}$ lead to contradictions. $b^{2}=a$ or $b^{2}=a^{5}$ imply |b|=12.
..........."

Why the $b^{2}=a^{2}, b^{2}=a^{4}$ lead to contradiction?
suppose $b^2=a^2.$ then from $bab^{-1}=a^{-1},$ we get $a^3b^{-1}=b^2ab^{-1}=ba^{-1}.$ thus: $a^{-3}=(bab^{-1})^3=ba^3b^{-1}=b^2a^{-1}=a.$ hence $a^4=e,$ which contradicts $|a|=6.$ do the same for the case $b^2=a^4.$