Can a group of order 98 have a subgroup of order 7?

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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.
 
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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