Question

Is it possible for a group G to contain a non-identity element of finite order and also an element of infinite order? If yes, illustrate with an example. If no, give a convincing explanation for why it is not possible.

Answer #1

Another way to show that in a free group, any non
identity element is of infinite order

Can there be an element of infinite order in a finite group?
Prove or disprove.

Give the definition of group. Show that the identity (or
neutral) element of G is
unique. Moreover, show that the inverse of an element g 2 G is
unique as well.

4. Let f : G→H be a group homomorphism. Suppose a∈G is an
element of finite order n.
(a) Prove that f(a) has finite order k, where k is a divisor of
n.
(b) If f is an isomorphism, prove that k=n.

Give an example of a nontrivial subgroup of a multiplicative
group R× = {x ∈ R|x ̸= 0}
(1) of finite order
(2) of infinite order
Can R× contain an element of order 7?

Prove that if (G, ·) is a finite group of even order, then
there always exists an element g∈G such that g ≠ 1 and
g2=1.

Let G be a group. g be an element of G. if
<g^2>=<g^4> show that order of g is finite.

True or False:
An infinite group must have an element of infinite order.
I know that the answer is false. Please give a detailed
explanation. Will gives thumbs up ASAP.

Let
G be a finite group and H a subgroup of G. Let a be an element of G
and aH = {ah : h is an element of H} be a left coset of H. If B is
an element of G as well show that aH and bH contain the same number
of elements in G.

2. Let a and b be elements of a group, G, whose identity element
is denoted by e. Prove that ab and ba have the same order. Show all
steps of proof.

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