Here is my attempt at proving that a group G of order 10 is isomorphic to the dihedral group D10.
By Cauchy's Theorem, there are elements r,s in G with orders 2,5 respectively. The subgroup < r > has index 2 in G; therefore it is normal. We look at the possibilities of what srs can equal.
- srs = e. Impossible since then r = e.
- srs = r, so sr = rs. Thus sr has order 10 (lcm of s, r) and G is cyclic.
- srs = r2. Then s = (r2)s(r4), so s2 = (r2)s(r)s(r4) = (r2)(r2)(r4) = r3 = e, false.
- srs = r3. Then s = (r3)s(r4), so s2 = (r3)[s(r2)]s(r4). From sr = (r3)s we have s(r2) = (r3)sr, so then s2 = rsrs(r4) = r8 = r(r3)(r4) = r8 = e, contradiction.
- srs = r4. Then < s, r > is, by definition of the dihedral group, isomorphic to D10 with order 10. Thus < s, r > is the whole of G, and G is isomorphic to D10.
Is this proof overcomplicated? My work in ruling out possibilities 2,3 seems especially inelegant. Is my conclusion in 5 correctly stated - particularly with my use of , by which I'm referring to the subgroup of G generated by s and r?
Thank you.
Can you use the Sylow Theorems (see https://en.m.wikipedia.org/wiki/Sylow_theorems)? If yes, then you can prove it much more easily:
Note that by the third theorem, there is exactly 1 sylow 5-group and either 1 or 5 sylow 2-groups. In the first case, by the third theorem, both the 2-group and the 5-group are normal subgroups. Further, note that they are generated by $r$ and $s$ (as the first theorem says that they are of order $2$ and $5$ respectively). Therefore, $r$ and $s$ commute ($rs = sr$) and $G$ is cyclic.
Alternatively, if there are 5 Sylow 2-groups, then there are 5 elements of order 2 (1 from each 2-group) and 5 elements of order 5 (from the 5-group). You should recognize this as $D_{10}$.
QED