1. ## Group Theory

a) If a group has order 12 , then G contains an element of order 6.

b) If a group has order 18, then G contains an element of order 3.

If it's true, prove it; if it's false find a counter example.

I know I should know this but for some reason can't get my head around it. Can someone give me a push in the right direction. Do I have to include the Sylow Theorems?

Thanx

2. Originally Posted by bex23
a) If a group has order 12 , then G contains an element of order 6.
This is not true for $A_4$ look here for the details.

b) If a group has order 18, then G contains an element of order 3.
Since $3|18$ by Cauchy's theorem there is an element of order 3.

3. [quote=ThePerfectHacker;177440]This is not true for $A_4$ look here for the details.

What about the group D_6. Doesn't that have 2 elements of order 6.

4. Originally Posted by bex23
What about the group D_6. Doesn't that have 2 elements of order 6.
That is true. Because $D_6$ is generated by $a=(123456)$ and $b=(26)(35)$. Thus. $D_6 = \{ a^ib^j | 0\leq i\leq 5, 0\leq j\leq 1\}$.
The elements $a^i b$ are all reflections and they have order 2.
While the order of $a^i$ is $6/\gcd(i,6)$.
To have order 6 we require $\gcd(i,6)=1$ thus $i=1,5$.
This means only $a$ and $a^5$ have order $6$.

But what is the problem? The above example is when it is true. But the example with $A_4$ is when it is false. Thus, this statement cannot be true i.e. order 12 subgroups has elements of order 6.