Prove that a group of even order must have an element of order 2.

Lewis Harvey

Lewis Harvey

Answered question

2021-01-06

Prove that a group of even order must have an element of order 2.

Answer & Explanation

joshyoung05M

joshyoung05M

Skilled2021-01-07Added 97 answers

To show: thereexist some x in G such that order of x is 2.
Let S={xG|x|>2}
Let lS
Since order of any element in group is equal to order of its inverse, thus |l1|=|l|>2
Hence, llS
Since, |l|>2 it implies that ll1. Hence every lement in S can be paired off with its inverse which is also on S. Thus, S otains even number of elements
GS={xG|x=eor|x=2}
Since , identity element is unique in a group, therefore G must have an element of order 2 in order to GS with even number of elements.
Therefore, a group of even order must have an element of order 2.

star233

star233

Skilled2023-05-28Added 403 answers

Step 1: Let G be a group of even order, and let |G| denote the order of G. By definition, the order of a group refers to the number of elements it contains.
Since G has even order, we can write |G|=2n, where n is a positive integer.
Step 2: Now, consider the set S={gG:gg1}. In other words, S consists of all the elements of G that are not equal to their own inverses.
Assume, for the sake of contradiction, that S is empty. This would mean that every element in G is its own inverse. Therefore, for any gG, we would have g=g1.
Next, let's consider the identity element e of G. By the definition of an identity element, e multiplied by any element gG results in g itself. Thus, eg=g.
Step 3: Now, since G is a group, it follows that g1G for any gG. Therefore, we can consider the product of an element and its inverse: gg1.
Using the fact that g=g1 for all gG (assuming S is empty), we can rewrite gg1 as ge.
This simplifies to g, since ge=g for any gG.
Thus, we have shown that g=g1 for all gG, which contradicts the assumption that S is empty.
Therefore, the set S cannot be empty. Hence, there exists an element hS such that hh1.
Step 4: Since h is not equal to its own inverse, we have h1h. This means that the order of h is greater than 1.
However, since |G|=2n, which is even, and h has an order greater than 1, it follows that the order of h must be even.
By Lagrange's theorem, the order of an element divides the order of the group. Since the order of h is even and divides the order of G, it must be a divisor of 2n.
The only possible divisors of 2n are 1, 2, n, and 2n.
Since h has an order greater than 1, we can conclude that the order of h cannot be 1 or 2n.
Therefore, the order of h must be either 2 or n. However, since hh1, it cannot have order n.
Hence, the order of h must be 2. Therefore, we have proven that a group of even order must have an element of order 2.
alenahelenash

alenahelenash

Expert2023-05-28Added 556 answers

To prove that a group of even order must have an element of order 2, we can proceed as follows:
Let G be a group of even order. By the Lagrange's theorem, the order of any element in G must divide the order of the group. Since the order of G is even, it can be expressed as |G|=2n, where n is a positive integer.
Now, suppose for contradiction that there exists no element in G of order 2. That means, for every element g in G, |g|2.
Consider the set S={g,g2:gG}. Since |g|2 for any g in G, we can conclude that |g2|2 as well. Therefore, each element in S has order greater than 2.
Note that every element in S is distinct. To see this, suppose gk=hm for some g,hG and positive integers k and m. Then, raising both sides to the power of 2, we have (gk)2=(hm)2, which implies g2k=h2m. Since g2k and h2m have order greater than 2, we can conclude that g2k=h2m if and only if gk=hm. Thus, the elements in S are distinct.
Now, let's count the number of elements in S. Since S consists of the elements g and g2 for every g in G, we have |S|=2|G|=4n.
On the other hand, note that S is a subset of G, and by Lagrange's theorem, the order of any subgroup of G must divide the order of G. Therefore, the order of S must divide |G|. However, we have |S|=4n, which does not divide 2n=|G|. This leads to a contradiction.
Hence, our assumption that there exists no element of order 2 in G is false. Therefore, a group of even order must have an element of order 2.

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