I could not verify the below paragraph in my textbook [Hungerford, p98].
Let , G is a group of order 12, and P is the normal (unique) sylow 3-subgroup. Hence G contains only two elements of order 3. If c is one of these, then or . Thus is a group of order 12 or 6. In either case there is of order 2 by Cauchy's theorem. Verify that |cd| = 6.
In my textbook, is such a group that has a normal sylow 3-subgroup.
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What is the normal sylow 3-subgroup of the above ?
unfortunately this is not always true, even if we assume the group is abelian. (for example, think about when ) we also need the orders to be coprime, which in this case:
however, it is true that there always exists an element of G of order this is a good problem which is worth thinking!
for the second part of aliceinwonderland's question: the Sylow 3-subgroup of is:
Yes , what was I thinking! But lucky my argument works because as you said.
I posted a solution a year ago, here via Sylow.however, it is true that there always exists an element of G of order this is a good problem which is worth thinking!
Thanks for all replies.
I have one last paragraph in my text that I could not understand regarding a nonabelian group of order 12.
"Let a = cd (in the above quote); then <a> is normal in G and |G/<a>| = 2. Hence there is an element such that .
Since G is nonabelian and is the only possibility.
....
There are six possibilities for lead to contradictions. or imply |b|=12.
..........."
Why the lead to contradiction?