If the group is simple, sylow is enough, otherwise i have no idea.
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I think you still use Sylow theory, except that there are two cases.
Let $G$ be a group of order 168. The number $n_7$ of Sylow 7-subgroups of $G$ must equal $1$ or $8$. If $n _7 = 1$, then there's a unique element of order 7. If $n_7 = 8$, there there are $8 \cdot 6 = 48$ elements of order 7.
manthanomen
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Is there no way to tell which is it? Thanks anyway – Kai Jan 26 '16 at 08:38
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Not unless you know more about the group. There could be non-simple groups of order 168 with $n_7 = 1$ and others with $n_7 = 8$. – manthanomen Jan 26 '16 at 08:48
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@manthanomen can you please explain why $8$ is possible and furthermore why multiply by $6$? – MrStormy83 Nov 09 '21 at 22:33