# #GroupTheory #TestBook

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Here we are given that the order of the group is 10 and a is the generator..

Hence  a10  =  e  where e is the identity element but we dont know about a8..Hence we have to write it in terms of a10 as that is a known result.

Let we have   y  =  a40   which can be written as :    a40  =   (a10)4   =  e4   =   e    ...(1)

Also ,        a40   =  (a8)5     =  e     [ Follows from (1) ]

Hence it means that  a8 is repeated 5 times in order to get the identity element e which is least number of times to do so.

Also from the corollary of the Lagranges' Theorem ,

Order of an element  divides the order of the order of the group (The actual theorem is order of a subgroup divides the order of the group)

Here 5 divides the order of group i,e, 10..

Hence order of  a8   =   5

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Plz explain this question
Let < G, * > be a finite group and let H1 and H2 be it's sub group as per lagrange theorem we know order of sub group divides order of group, my doubt is, is it possible that we can have multiple subgroup of same cardinality for a particular group ? $\left | H1 \right |=\left | H2 \right |$