Let $G$ be a group of order $56$. Then which of the following are true Let $G$ be a group of order $56$. Then which of the following are true   


*

*All $7$-sylow subgroups of $G$ are normal   

*All $2$-Sylow Subgroups of $G$ are normal   

*Either a $7$-Sylow subgroup or a $2$-Sylow subgroup of $G$ is normal    

*There is a proper normal subgroup of $G$.  


How would I able to solve this problem and which theorem(s) would be required to solve this? Thanks for your time.
 A: First, note that statement $4$ must always be true, since the identity subgroup is always normal.  (I suspect they accidentally omitted the word 'nontrivial'.)
Second, note that it is certainly possible that $G$ contains both a unique Sylow $7$-subgroup and a unique Sylow $2$-subgroup, evidenced by the cyclic group of order $56$.  So statement $3$ is false if you interpret the 'or' as exclusive (which, grammatically, is implied by the 'either').  Otherwise, we proceed as follows.

Suppose that no Sylow $7$-subgroup of $G$ is normal.  Since $7^1$ is the highest power of $7$ to divide $56$, every $P\in \text{Syl}_7(G)$ has order $7$.  Taking another $Q\in \text{Syl}_7(G)$, we have that $P\cap Q=1$.  Then the Sylow $7$-subgroups (excluding the identity) make up $6\times n_7$ distinct elements of $G$, all of which have order $7$.  We have assumed that $n_p\not= 1$, so since $n_7\equiv 1 \mod 7$ and $1+6\times(7m+1)>56$ for any $m\geq 2$, we find that $n_7=8$.
That leaves us with $56-(8\times 6)=8$ elements with which we can form Sylow $2$-subgroups.  Since $2^3=8$ is the highest power of $2$ dividing $56$, this is exactly enough for one Sylow $2$-subgroup, which must then be normal.

So that proves the implication $n_7\not= 1 \Rightarrow n_2= 1$, which we can extend to $(n_7=1)\vee (n_2=1)$.  Thus $3$ is true if we take the "or" to be inclusive.  (Of course, this implies statement $4$ even with a nontriviality condition.)
We still need to examine statements $1$ and $2$, however.  Both statements are false, so we need to construct counterexamples for each.  For statement $1$,

Assume that $n_7\not= 1$ and let $P\in \text{Syl}_7(G)$ and $Q\in \text{Syl}_2(G)$.  Then $G=Q\rtimes P$ and $P$ acts on $Q$ by conjugation.  The orbit of any element of $Q$ has size $7$ (since the semidirect product is nontrivial), so since $Q$ has size $8$, all nonidentity $q\in Q$ are in the same orbit.  If we take $q_0$ to be the element of order $2$ in $Q$, we see that $o(q_0)=o(q^p)$.  Thus $Q\cong \mathbb{Z_2}\times \mathbb{Z_2} \times \mathbb{Z_2}$.
Since $|GL_3(\mathbb{F}_2)|=(2^3-2^0)(2^3-2^1)(2^3-2^2)=7\times 6 \times 4$ is divisible by 7, $\text{Aut}(\mathbb{Z}_2^3)$ does indeed have an element of order $7$, so we can in fact form such a semidirect product.  Thus there exists a group of order $56$ which does not have normal Sylow $7$-subgroups.

Constructing a counterexample for statement $2$ is a bit easier.

Let $P=\mathbb{Z}_8=\langle a \rangle$ and $Q=\mathbb{Z}_7$, so that $\text{Aut}(Q)\cong \mathbb{Z}_6=\langle \theta \rangle$.  Let $G$ be the semidirect product $Q\rtimes_\phi P$ defined by $a\mapsto \theta^3$.  Then $G$ is a group of order $56$ and $P$ is not normal in $G$.

In summary, to provide a complete answer to that question, you need Sylow theorems for statement $3$ and knowledge of nontrivial semidirect products to disprove statements $1$ and $2$.
A: Let $G$ be a group of order $56 = 2^3\times7$. Let $n_2$ be the number of Sylow $2$-
subgroups in $G$ and $n_7$ be the number of Sylow $7-$subgroups in $G$. We will show that
$n_2 = 1$ or $n_7 = 1.$ Using the third Sylow theorem, $n_2 = 1$ or $7$, and $n_7 = 1$ or $8.$
If $\,n_7 = 1\,$ then we are done. So assume that $n_7 = 8$. Then there is one element
of order $1$ and $48 = 8\times 6$ elements of order $7$. One Sylow $2-$subgroup contributes $7$
new elements (whose orders divide $8 = 2^3 $). Therefore, having more than one Sylow
$2-$subgroup means there are more than 56 elements in $G$, which contradicts the fact
that $|G| = 56$. Hence, there is only one Sylow 2-subgroup.
A: Let $G$ be a group with $|G| = 56 = 2^3
· 7$ so that the number of Sylow $2$-subgroups of $G$ is $1$ or $7$
and the number of Sylow $7-$subgroups is $1$ or $8$. Suppose that $G$ does not have a normal Sylow $7-$subgroup.
Then there are $8$ Sylow $7-$subgroups. Now the intersection of two distinct Sylow $7-$subgroups must be trivial
since every nontrivial element of a group of order $7$ generates the group. It is clear that the intersection of
any Sylow $2-$subgroup with any Sylow $7-$subgroup is also trivial. Therefore, $G$ contains $48$ elements of order
$7$. This leaves $56 − 48 = 8$ elements not of order $7$, so that we have exactly one Sylow $2-$Subgroup. Hence,
$G$ has a normal Sylow p-subgroup for $p = 2$ or $7$.
