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I'm not sure my proof is correct because I don't use the fact that the group is simple anywhere...

Since $168=2^3 \cdot 3 \cdot 7$, there exists a subgroup of order 7. Let $n_7$ be the number of Sylow 7-subgroups. Since $n_7 | 24$ and $n_7 \equiv 1$ mod $7$, $n_7 = 8$. Since every group of prime order is cyclic, there exist 8 elements of order 7.

I think that I still need to prove that there are no other elements of order 7 and that's where the fact that the group is simple would come in.

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  • $\begingroup$ I do not understand " Since every group of prime order is cyclic, there exist 8 elements of order 7".... any element of order $7$ has to be in sylow $7$ group so only thing you have to do is correctly count exact number of sylow $7$ subgroups $\endgroup$
    – user87543
    Mar 21 '14 at 7:12
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The $8$ Sylow subgroups of order $7$ each contain $6$ elements of order $7$ together with the identity. Since an element of order $7$ generates a group of order $7$ these elements are all distinct. There are $8 \times 6=48$ elements of order $7$.

You use the fact that the group is simple in asserting that there are $8$ subgroups of order $7$. It is always possible to have $1$ subgroup of a given order (so long as it is a factor of the group order e.g. the cyclic group of order $168$) - but a single subgroup of order $7$ would be normal, and the group would not be simple.

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  • $\begingroup$ How do you know that every element in the Sylow subgroup except the identity has order 7? $\endgroup$
    – Rebekah
    Mar 21 '14 at 16:13
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    $\begingroup$ @user124104 There is only one kind of group of order a prime $p$ - the cyclic group. In any group the orders of the elements divide the order of the group - so the orders must be $1$ or $p$. Only the identity has order $1$, so the remaining $p-1$ elements have order $p$. I've written this in terms of a general prime because you will be looking at Sylow subgroups for different primes, I'm sure. When the order is $p^2$ there are two possible group structures - but all the Sylow $p$-subgroups fro a given $p$ are conjugate, and therefore have the same structure. $\endgroup$ Mar 21 '14 at 16:27
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Making use of the theorem-

Let $G$ be a group of order $p^{\alpha}m$,where $p$ is a prime,$m\ge 1$ and $p$ does not divide $m$,then the number of sylow $p-$subgroup of $G$ is of the form $1+kp$ i.e,$$n_p \equiv 1(modp)$$.Further,$n_p$ is the index in $G$ of the normalizer $N_G(P)$ for any Sylow $p-$ subgroup of $P$,hence,$n_p$ divides $m$.

Since,$\vert G\vert=168=2^3\cdot 3\cdot 7$;

$n_7=1+7k$ for some integer $k$,and $n_7$ divides $2^3\cdot 3=24$,this restricts the only possibility for $k$ to be $1$ only. So, $n_7=1+7\cdot 1=8$.

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168=2^3*3*7 and we can find element of order 7 only in the subgroup of order 7. so number of sylow subgroups of order 7 ,1+7k ie 8 because group is simple and unique sylow subgroup is normal .no of elements of order 7 in a one group of order 7(prime order=cyclic)=phi(7)=6 .so total no of elements of order 7 in the whole group (of order 168) ie. on whole 8 subgroups =8*6=48

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168=8⋅3⋅7 the elements of sylow 7-groups of grop of order 168 the order of 7 subgroups is 7. number of elements between 0-168 which is 1 mod 7 are 2,8,15,22,29... since 1 and 8 are the only positive integers congruent to 1 mod 7 and divides 168, the number of sylow 7-subgroups is 8. then there are 8x6=48 elements of order 7 since identity is of order 1 and common in all the 8 sylow 7-subgroups

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