(a) Prove that there are three abelian groups of order 56.
(b) Prove that ever group of order 56 has either a normal Sylow 2-subgroup or a normal Sylow 7-subgroup.
(c) Assume there is a normal Sylow 7-subgroup, and construct the following groups with the Sylow 2-subgroup specified:
$~~~~~$one group when $S ≅ Z_2 \times Z_2 \times Z_2$,
$~~~~~$two nonisomorphic groups when $S ≅ Z_4 \times Z_2$,
$~~~~~$one group when $S ≅ Z_8$,
$~~~~~$two nonisomorphic groups when $S ≅ Q_8$,
$~~~~~$three nonisomorphic groups when $S ≅ D_8$.
[For a particular $S$, two groups are not isomorphic if the kernels of the maps from $S$ to $Aut(Z_7)$ are not isomorphic.]
(d) Let $G$ be a group of order 56 with a nonnormal Sylow 7-subgroup. Prove that if $S$ is the Sylow 2-subgroup of $G$, then $S ≅ Z_2 \times Z_2 \times Z_2$.
(e) Prove that there is a unique group of order 56 with a nonnormal Sylow 7-subgroup. [For existence, use the fact that $|GL_3(\mathbb{F}_2)|=168$; for uniqueness use exercise 6.]
Proof: (a) By partitioning 56 by its elementary divisors, we easily obtain the groups $Z_8 \times Z_7$, $Z_4 \times Z_2 \times Z_7$, and $Z_2 \times Z_2 \times Z_2 \times Z_7$.
(b) If both were nonnormal, then by Sylow's theorems, we could only have $n_2 = 7$ and $n_7 = 8$. Add up the nonidentity elements of the Sylow 7-subgroups, all the elements of a Sylow-2 subgroup, and at least one distinct element of another Sylow 2-subgroup to obtain $8*6 + 8 + 1 = 57 > 56 = |G|$.
(c) For the first three cases, note that a homomorphism defined on an abelian group's factor's generators extends uniquely to a group homomorphism since up to multiplicative rearrangement of the terms, there is a unique way of writing the elements of the group. This covers the gist of the proof, but further details may be obtained here as it applies to elementary abelian groups. In any case, this allows us only to define a homomorphism merely by defining it for the generators, and thus allows us to define $S$'s potential actions on $H$ as part of the semidirect product by defining the action for the generators of $S$.
Since the Sylow 7-subgroup $H$ is cyclic, we have $Aut(H) ≅ Z_6$ by its abelianness and order, so that the only nontrivial homomorphism from $S$ of order 8 into $Aut(H)$ has for image the unique subgroup of order 2, which contains the identity automorphism and the inverting automorphism. Thus, we might have $s \cdot h = h^{-1}$ for all $h∈H$ and for $s∈\{(0,0,1),(0,1,0),(1,0,0)\}$.
For the next case, it is similar, as the only available homomorphisms are those mapping $S$ to the inversion subgroup of $Aut(H)$. Define two actions of $S$ on $H$ thus: $s \cdot h = h^{-1}$ for $s∈\{(1,0),(0,1)\}$, or $s \cdot h = h^{-1}$ for $s∈\{(1,0)\}$ and $s \cdot h = h$ for $s∈\{(0,1)\}$. The former utilizes a homomorphism from $S$ into $Aut(H)$ with a kernel of order 4 and containing $(1,1)$ of order 4, so the kernel is isomorphic to $Z_4$. The latter produces a homomorphism with kernel of order 4 containing two normal, trivially intersecting subgroups of order 2 $\langle (2,0) \rangle$ and $\langle (0,1) \rangle$ so that it is isomorphic to $Z_2 \times Z_2$. Thus these two semidirect products afforded by these actions are distinct by the footnote.
For the third case, we must have $s \cdot h = h^{-1}$ for $s∈\{x\}$ where $x$ generates $S$.
For the fourth case, we can have $s \cdot h = h^{-1}$ for $s∈\{i,j\}$, and compare it with the trivial homomorphism producing $H \times Q_8$. The former has a nontrivial kernel, so these two are distinct.
For the fifth case, we can have $s \cdot h = h^{-1}$ for $s∈\{r,s\}$, and $s \cdot h = h^{-1}$ for $s∈\{s\}$ and $s \cdot h = h$ for $\{r\}$, as well as the trivial homomorphism producing $H \times D_8$. The first has a kernel of $\langle r^2, sr \rangle$, the second $\langle r \rangle$, and the third the identity, which are easily checked to be isomorphic to $Z_2 \times Z_2$, $Z_4$, and the identity group respectively, so that these are all distinct semidirect products.
(d) We have $G ≅ S \rtimes_φ H$ as usual, where $H$ is the nonnormal Sylow-7 subgroup and $S$ must thus be normal. Since $H$ is nonnormal, the homomorphism $φ$ must not be trivial, so that the kernel of $H$ is the identity. Let $H$ act on $S$ by conjugation: If $h^a \cdot s = h^b \cdot s$ for any nonidentity $h$ and $s$, then $h^a = h^b$, so that every element of $S$ can be expressed as $h^ash^{-a}$ for some integer $a$ producing a distinct element of $H$. These all have the same order as $s$, so that all of the elements of $S$ have the same order, and by Cauchy's theorem there exists an element of order 2 so that $S$ must be isomorphic to $Z_2 \times Z_2 \times Z_2$.
(e) Since $|Aut(S)| = |GL_3(\mathbb{F}_{19})| = 168 = 2^3*3*7$, we have that all the possible nontrivial homomorphisms from $H$ into $Aut(S)$ are onto the Sylow 7-subgroups thereof (map generator to generator for homomorphism), which are established to be conjugate. By this fact, the cyclicity of $H$, and exercise 6, we have the sole semidirect product obtained. $\square$
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