I'm trying to solve the following problem using the Sylow theorems:
Determine all groups of order $\leq 10$ up to isomorphism.
I know that in particular I have to use the fact that any p-subgroup is contained in a p-Sylow subgroup, but I don't see how to use that to find the groups, let alone show there are no others. Thanks for any help!
On
Here is the case $|G| = 6$ (it's fairly easy, but you can see the ideas at work).
From Sylow theory, we know we have a subgroup of order $2$, and a subgroup of order $3$.
Any subgroup of order $3$ is normal (being of index $2$), and thus is unique (since all Sylow $3$-subgroups are conjugate).
Straight off the bat, that gives us $4$ distinct elements of $G$:
$\{e,a,a^2,b\}$ where $o(a) = 3$ and $o(b) = 2$.
Let's call $P = \langle a\rangle$, and $Q = \langle b\rangle$.
Since $\gcd(2,3) = 1$, we know that $G = PQ$.
So we want to know if $Q \lhd G$. If so, then $G \cong P \times Q$, and the Chinese Remainder Theorem tells us $G \cong C_6$ (a cyclic group of order $6$).
Now the number of Sylow $2$-subgroups is odd (congruent to $1$ mod $2$), and a divisor of $6/2 = 3$. So $1$ and $3$ are the only possibilities. If the number of $2$-sylow subgroups is $1$, we are done (see above).
Otherwise, we have $3$ Sylow $2$-subgroups, and none of them are normal.
In this case, $G$ is the semi-direct product of $P$ and $Q$. In other words, $Q$ acts on $P$ by conjugation.
Since conjugation by the non-identity element $b \in Q$ yields an automorphism of $P$, it must send $a$ to either $a$ or $a^{-1} = a^2$.
If it sends $a \to a$, this means conjugation by $b$ fixes (every element of) $P$, and since conjugation by any element of $P$ also fixes every element of $P$, this means that the subgroup:
$H = \{g \in G: gxg^{-1} = x\ \forall x \in G \}$ has at least $4$ elements, and thus is all of $G$.
(Why can we conclude that if conjugation by $b$ fixes $a$, that $b$ commutes with everything? Because $G = PQ$, so everything is a product of $a's$ and $b's$. Similar reasoning holds for conjugation by $a$).
If $H = G$, then $G$ is abelian, and $Q$ would be normal. But we assumed $Q$ non-normal.
Thus we have to have $bab^{-1} = a^{-1}$.
We can see that the remaining two elements of $G$ we had not yet explicitly identified are $ab$ and $ba$. From our relation: $bab^{-1} = bab = a^{-1}$, we see that:
$ba = a^{-1}b = a^2b$.
In other words; $G = \langle a,b: a^3 = b^2 = e, ba = a^2b\}$, so $G$ is isomorphic to $D_3 \cong S_3$, which we can regard as either the symmetry group of the equilateral triangle, or the full symmetric group on $3$ letters.
There are more elementary ways to show the above-the idea of this is to indicate how the same techniques can work on groups of larger orders, where we might have more possibilities.
Suppose that $|G|=8$ (all other cases are easier).
It is clear from structure theorem for finitely generated abelian groups, that thete are only three abelian groups (up to isomorphism): $\mathbb{Z}_8$, $\mathbb{Z}_4\oplus \mathbb{Z}_2$ and $\mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}_2$.
Suppose that $G$ is a non-abelian group. Then all elements from $G\setminus \{e\}$ have order 2 ore 4. If all elements from $G\setminus \{e\}$ have order 2 then $G$ is abelian (it is an easy exercise). So there is element $x$ of order 4.
First case. Suppose that there is an element $y$ of order 2, such that $y\notin \left \langle x \right \rangle_4$. In this case all elements $$x^ky^l, k=0,1,2,3; l=0,1$$ are different. Since $(G:\left \langle x \right \rangle_4) = 2$ subgroup $\langle x \rangle_4$ is a normal subgroup in $G$. Then $y^{-1}xy=a^k$, where $k=1$ or $k=3$. If $k=1$ then $G$ is an abelian group, so $k=3$. This is $\textbf{D}_4$.
Second case. Suppose that if $y\notin \left \langle x \right \rangle_4$ then $\textrm{ord}(y)=4$. Then $y^2\in \left \langle x \right \rangle_4$ and this gives us that $y^2 = x^2$. Then $z=x^2=y^2\in Z(G)$. Similat to the first case one can show that $y^{-1}xy = x^3 = zx$. Now if we put $z=-1, x=i, y=j$ and $xy = k$ we will get $\mathbb{Q}_8$.
P.S. I know, that my English is terrible.