[代數] order of a group

[ssss50201 (ssss)]

需要幫忙確認做法正確~~ Thanks :)

If the finite group G contains a subgroup of order 7
but no element (other than the identity) is its own inverse,
then the order of G could be
A.27 B.28 C.35 D.37 E.42


By Larange, A,D 刪掉

no element (other than the identity) is its own inverse
這句話其實有點不懂
意思是 a =! a' for all a belongs to S\{identity} 嗎？
ps.這邊我讓S表示subgroup of G. a'表示inverse of a

如果意思是這樣，從這個條件可以得知G has no subgroup of order 2
(因為如果a = a', 那aa=aa'=identity, subgroup of order 2)

然後B,E就被刪掉


答案剩下C


ps. 其實關於order of an element & order of a group我有點困惑
order of a group 是指cardinality
order of an element 是指最小的m, st a^m=identity.
但有時候order of an element的定義似乎又可以用來定義order of a group
是不是我弄混了什麼事呢？

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◆ From: 108.3.154.49
推 APM99       :最後倒數第二行  是因為那個group是由那個element
→ APM99       :生成的.
→ ssss50201   :所以 當order of element被用來當作是order of group
→ ssss50201   :時，前提是該element生成了G囉？
→ ssss50201   :如果是的話那我的做法就不對了@_@