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# Argue why every odd order subgroup has to be a subgroup of the cyclic odd order subgroup K of index 2, and by the divisibility argument it still could be a subgroup of G and not to be contained entirely in K.

Please see the attached file first.

I did the proof, but it's weak, since I can't find a way to argue why every odd order subgroup has to be a subgroup of the cyclic odd order subgroup K of index 2, and by the divisibility argument it still could be a subgroup of G and not to be contained entirely in K.

#### Solution Preview

You did a good proof, the only thing you did not explain is that why every
subgroup of odd order must be contained in the subgroup of order K.
Here is my proof.

Suppose K is the cyclic subgroup of order k in G and |G|=2k. We can find
an element h with order 2 in G, then h is not in K. (Because if h is in K, then
<h> is a subgroup of K, then |<h>|=2 divides |K|=k, but k is odd, 2 does not
divide k, this is a contradiction.) Then G ...

#### Solution Summary

This solution is comprised of a detailed explanation to argue why every odd order subgroup has to be a subgroup of the cyclic odd order subgroup K of index 2, and by the divisibility argument it still could be a subgroup of G and not to be contained entirely in K.

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