Hello,

For the dihedral groups we are using the $D_{2n}$ notation where $n$ is the number of vertices and $2n$ is the order of the group.

I am trying to find a counter example to "Let $G$ be a group of order $2^m o$ (where $o$ is some odd number) find an element of order $2^m$ in $G$"

The order of $D_2*6$ is $( 2^2)*3$ so $D_{12}$ should have an element of order 4.

I can discard all elements of form $sr^i$ (for some $0 \leq i < 6$) as each has order 2 (by $(sr^i)^2=srr^{-1}s=e$ if you like) so I am left with only $\{1 , r , r^2 , r^3 , r^4 , r^5\}$ and $| 1 |=1, | r | = 6, | r^2 | = 3, | r^3 | = 2, | r^5 | = 6$.

Now I think that I am out of elements but I have not found an element of $D_{12}$ which has order 4.

Is this correct? Or have I missed possibly many elements?