# Group of order 63 has subgroup of order 21

Is there any way to prove that a group of order 63 has a subgroup of order 21 just using Cauchy's Theorem (i.e., without Sylow)?

We know that there is an element $x$ of order $7$ and an element $y$ of $3$. Can we conclude that $xy$ has order 21, since $7$ and $3$ are coprime?

• $xy$ need not have order 21, however, you will still have a subgroup of order $21$. – Justin Benfield Nov 11 '16 at 7:01
• If $x$ and $y$ commuted, then it would follow that $xy$ has order 21. But this is not true in general. For example, consider $(1234567)$ and $(765)$ in $S_7$, whose product has order 5. In fact, there is a nonabelian (in particular noncyclic) group of order 21, and your elements $x$ and $y$ could generate such a subgroup. (You can however show using Sylow's theorem that there exists some element of order 21 in a group of order 63; I haven't tried doing this without Sylow.) – Ravi Fernando Nov 11 '16 at 7:14
• Problem is discussed at artofproblemsolving.com/community/… but using Sylow, I think. Also done, using Sylow, in Chapter 4 of math.utah.edu/~malone/prelims/algebra.pdf – Gerry Myerson Nov 11 '16 at 8:22
• Hint: Looking at the normalizer of a subgroup of order 7 one gets that there are 8 elements of order 3, on which the group acts by conjugation. – j.p. Nov 11 '16 at 8:43
• I should have written "8 elements of order a multiple of 3" instead in my last comment. – j.p. Nov 11 '16 at 9:45

By Cauchy you have a subgroup $S$ of order 7. Its normalizer $N :=N_G(S)$ has order 7, 21 or 63. If 21, then you are done. If 63, then $N$ is normal, so take the preimage of a subgroup of order 3 of $G/N$, which exists by Cauchy. If 7, then $N$ has (counting itself) 9 conjugates, which intersect trivially, as $N$ has prime order. This gives $54=9\cdot 6$ elements of order 7. Having one element of order 1, this leaves $8=63-54-1$ elements unaccounted for, some of which have order 3 by Cauchy. $N$ acts by conjugation on the set $T$ of elements of $G$ of order 3. Elements of order 3 come in pairs (paired with their inverses) so $T$ has even order at most 8, i.e., 2, 4, 6 or 8. Hence $N$ has a fixed point $t$, which it centralizes (as the action is by conjugation), so $N$ and $t$ generate a subgroup of order 21 (or you observe that the centralizing element contradicts the normalizer having order 7, if you prefer to end the proof this way).
• Thanks for the response. How do we know that $N$ has $9$ conjugates? And how do we know that $N$ fixes some point just from the fact that $T$ has even order of at most $8$? – CuriousKid7 Dec 20 '16 at 21:41
• The number of conjugates of a subgroup $U\le G$ equals the index of its normalizer $N$ in $G$: The conjugate of a subgroup is again a subgroup, hence the group acts on the set of its subgroups. The orbit of this action containing $U$ is the set of conjugates of $U$. The point-stabilizer of the point $U$ with respect to this action is its normalizer $N$, and the length of an orbit is the index of the point-stabilizer of any point in this orbit. – j.p. Dec 21 '16 at 6:49
• The possible lengths of orbits of any action of a (finite) group $N$ are the divisors of the cardinality of $N$ (all divisors that equal the index of some subgroup of $N$), here 1 or 7. An orbit of length 1 is a fixed point. $T$ is the (disjoint) union of orbits. – j.p. Dec 21 '16 at 6:56
• Thank you. The proof makes sense. Just one more thing: You mentioned that the existence of $t$ contradicts that $|N|=7$. Why is this? – CuriousKid7 Dec 21 '16 at 23:59
• You're welcome. If $N$ centralizes $t$, then $t$ centralizes (and hence normalizes) $N=S$, so $t\in N$, leading to a contradiction. – j.p. Dec 22 '16 at 2:56