阶为2p的群是循环群或二面体群，为何p=2时情况平凡？

Great question! Let's unpack why the (p=2) scenario is considered "trivial" here:

When (p=2), the group (G) has order (2 \times 2 = 4). A core result in introductory group theory tells us there are only two distinct isomorphism classes of groups with order 4:

• The cyclic group (\mathbb{Z}_4) (this directly fits the "cyclic group" category in your original statement)
• The Klein four-group (V_4) (which is isomorphic to the dihedral group (D_2), matching the "dihedral group" part of the conclusion)

Since every group of order 4 must be one of these two, and both perfectly satisfy the theorem's claim ("either cyclic or dihedral"), we don't need to use the more complex arguments required for odd primes (like Sylow theorems, analyzing element orders, or constructing group actions). The classification of order-4 groups is a standard, straightforward result—so it's labeled "trivial" because it doesn't demand the non-trivial reasoning needed when (p) is an odd prime.

In short: we already know all possible groups of order 4, and they exactly cover the two cases the theorem describes. No extra work needed!

内容的提问来源于stack exchange，提问作者newhere