let G be any group. show that an element in G and it's inverse have the same order.

Expert Answer to "let G be any group. show that an..."

2022-05-03

xleb123

Skilled2023-05-04Added 181 answers

To show that an element

g

in group

G

and its inverse

g^{- 1}

have the same order, we need to prove that

| g | = | g^{- 1} |

, where

| g |

denotes the order of

g

.
Let

| g | = n

. This means that

g^{n} = e

, where

e

is the identity element of

G

. We want to show that

| g^{- 1} | = n

.
First, we note that

(g^{- 1})^{n} = (g^{n})^{- 1} = e^{- 1} = e

. This shows that the order of

g^{- 1}

is at most

n

.
Now, suppose for the sake of contradiction that the order of

g^{- 1}

is less than

n

. Let

m

be the order of

g^{- 1}

, so that

(g^{- 1})^{m} = e

. Since

m < n

, we have

g^{n m} = e

. However, since

g^{n} = e

, we also have

g^{n m} = (g^{n})^{m} = e^{m} = e

. This implies that

g^{n m} = e

, which contradicts the assumption that

| g | = n

. Therefore, we must have

| g^{- 1} | = n

, and the proof is complete.

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