# 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

1. All $7$-sylow subgroups of $G$ are normal
2. All $2$-Sylow Subgroups of $G$ are normal
3. Either a $7$-Sylow subgroup or a $2$-Sylow subgroup of $G$ is normal
4. 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.

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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 $|GL_3(\mathbb{F}_2)|=(2^3-1)(2^3-2)(2^3-4)=7\times 6 \times 4$ 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 $2$.

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identity subgroup is improper normal subgroup of G – rst Jan 19 '14 at 12:02
@rst Would you mind explaining that comment? I don't understand what you're trying to say. – Alexander Gruber Jan 20 '14 at 15:43
See option $4$ of the question. you are saying that option $4$ is always true as identity subgroup is always normal.But identity subgroup is improper normal subgroup of G – rst Jan 22 '14 at 15:46
@rst What is an "improper" subgroup? If you mean to say that the identity subgroup is not a proper subgroup of $G$, this is only true when $G$ is the group with $1$ element. – Alexander Gruber Jan 22 '14 at 17:18

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.

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sorry dear sir? I can not swim does it mean that I can not drive a car? – La Belle Noiseuse Dec 18 '12 at 9:04
The answer above has some logical flaws: as you say, if$\,n_7=8\,$ then there are $\,8\cdot 6=48\,$ elements of order 7 in the group, so the remaining $\,56-48=8\,$ elements must be the unit and the $\,7\,$ elements of the then unique $\,2$-Sylow subgroup, which is thus normal, and we're done! OTOH...nothing, since if $\,n_2=7\,$ then we cannot be sure how many elements of order a power of $\,2\,$ are there as the intersections between different Sylow-$2\,$ groups may be non-trivial...In fact, it happens to be there are groups of order $\,56\,$ with more than one $\,2-$Sylow subgroups. – DonAntonio Dec 18 '12 at 11:51
@Kuttus It seems that you drive the car as well as you can swim :)) – user26857 Dec 22 '12 at 2:06