Sylow's Theorems question

Hi
Why must a group of order 98 contain a subgroup of order 7?
I would think that Sylow's 1st theorem implies there exists at least one Sylow-7-subgroup of order 49 and at least one Sylow-2-subgroup of order 2 (since 98=2x7x7).
Thanks

Ray Veldkamp
 
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7 is prime, 7 divides 98, hence by Cauchy's theorem there is an element of order 7. The subgroup generated by this element is of order 7.
 

mathwonk

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or because the stronger version of sylows theorems say there is always a subgroup of any prime power order that divides the order of the group, not just the maximal prime power order.
 

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