Exhibit all Sylow 2- and Sylow 3-subgroups of and .

The Sylow 2-subgroups of have order 4. By Sylow’s Theorem, the number of Sylow 2-subgroups is odd and divides 3; hence . Note that

are distinct subgroups of order 4. Thus , and we have exhausted the list of Sylow 2-subgroups of .

The Sylow 3-subgroups of have order 3, hence are cyclic. Note that any element of the form sr^a$ has order 2, and so is not in a Sylow 3-subgroup. Thus the Sylow 3-subgroups are contained in ; this group is isomorphic to , and thus has a unique subgroup of order 3, namely .

The Sylow 2-subgroups of have order 4. Now by Sylow’s Theorem, the number of Sylow 2-subgroups is odd and divides 9; thus . Note that the following are subgroups of order 4.

Moreover, these groups are all distinct since for any pair, some generator of one is not contained in the other. Thus and we have exhausted the Sylow 2-subgroups of .

The Sylow 3-subgroups of have order 9. By Sylow’s Theorem, the number of Sylow 3-subgroups divides 4 and is congruent to 1 mod 3. Thus . Recall that has one element of order 1, three elements of order 2, and two elements of order 3, and that the order of is the least common multiple of the orders of and . So every element of has order 1, 2, 3, or 6, and moreover we can count that there are 15 elements of order 2, 8 of order 3, and 12 of order 6. Since every element in a Sylow 3-subgroup must have order 3, the elements of order 3 together with the identity comprise the only possible Sylow 3-subgroup. So .

Now is a subgroup of order 9, thus is the unique Sylow 3-subgroup of .

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## Comments

How can you conclude that in the last paragraph? It can be , but after checking the actual subgroup we can deduce that it is unique.

You’re right. Thanks!