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I tried getting $n_{2}$ and $n_{7}$ , denote the number of Sylow-2-Subgroups and Sylow-7-Subgroups respectively.

I got two cases for $n_{2} = 1 , 7 $ and for $n_{7} = 1 , 8$ , i noticed that if $n_{2} = 1$ and $n_{7}= 1$ ,then we are done since they are unique subgroups and hence will be normal in $G$.

Next how to proceed with the cases of $n_{2} = 7$ and $n_7 = 8$.

Please help?

2 Answers 2

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Count the total number of elements. Suppose neither $n_2$ nor $n_7$ is equal to $1$. Then $n_7=8$, which makes $8\cdot 6$ elements which belong to a $7$-Sylow subgroup, and to no other Sylow subgroup.

On the other hand, since $n_2=7$, each $2$-Sylow subgroup contains at least $4$ elements which belong to no other Sylow subgroup. In all $7\cdot 4$ elements.

Summing up, the group would contain at least $48+28+1$ elements. This is more than $56$, if I'm not mistaken.

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    For the case of $n_{7} = 8$ i think you are removing the trivial cases of identity and the group itself thus remaining with $6$ elements in each Sylow -7-subgroup thus giving $8.6 = 48$ elements belonging to all $8$ Sylow 7 subgroups.Is this correct? – 2017-01-28
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    Perfectly correct! – 2017-01-28
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    Now for $n_{2}=7$ we must have $5$ elements left , but why there are atleast $4$ mentioned ? – 2017-01-28
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    I considered the worst case: all $2$-Sylow all intersect in the *same* subgroup of order $4$: there remains $4$ elements in each $2$-Sylow. – 2017-01-28
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    all 2-Sylow all intersect in the same subgroup of order 4? – 2017-01-28
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    Summing up, the group would contain at least 48+28+1 elements. I think the group here is the union of both Sylow 2 and 7 subgroups whose order = number of elements exceeds the order of the original group , but i dont understand that we would have produced a contradiction even when we would have considered the identity and the group itself , then why are we removing them from counting?.. – 2017-01-28
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    The identity and the group itself are not Sylow subgroups! – 2017-01-28
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We have $o(G)=56=2^3.7$

Possible no. of sylow 2 subgroup $n_2=1+2k|7$ i.e either $1$ or $7$.

Similarly possible no. of sylow $7$ subgroup $n_7=1+7k|8$ i.e either $1$ or $8$.If there are $1$ sylow $7$ subgroup,then we are done because unique sylow-$p$ subgroups are normal.Suppose there are $8$ distinct sylow-$7$ subgroup say, $p_1,p_2\cdots p_8$.

Now $$|p_i\cap p_j||_{|p_i|or|p_j|}$$ $$|p_i\cap p_j|=1;i\neq j$$ $$|p_1\cup p_2\cdots\cup p_8|=56-7=49$$ i.e there are $48 $ elements of order $7$.We know that order of sylow- $2$ subgroup of $G$ is $8$.If there are $8$ sylow-$7$ subgroup then, there can only $1$ sylow-$2$ subgroup is possible which will be normal.

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    How the pairwise distinct intersection order is $1$? . – 2017-01-28
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    Because $ p_i\cap p_j$ are subgroups of $p_i$ and $p_j$.So, $| p_i\cap p_j|$ is either $1$ or $7$ (by Lagrange's theorem) but $7$ is not possible because $p_i$'s are distinct sylow-$7$ subgroup of $G$. – 2017-01-28