Question

Suppose G is a group and H is a normal subgroup of G. Prove or disprove ass appropirate. If G is cyclic, then G/H is cyclic. Definition: A subgroup H

Abstract algebra
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asked 2021-01-25
Suppose G is a group and H is a normal subgroup of G. Prove or disprove ass appropirate. If G is cyclic, then \(\displaystyle\frac{{G}}{{H}}\) is cyclic.
Definition: A subgroup H of a group is said to be a normal subgroup of G it for all \(\displaystyle{a}\in{G}\), aH = Ha
Definition: Suppose G is group, and H a normal subgruop og G. THe froup consisting of the set \(\displaystyle\frac{{G}}{{H}}\) with operation defined by (aH)(bH)-(ab)H is called the quotient of G by H.

Answers (1)

2021-01-26
Let G be a group, and H be a normal subgroup of G. Assume that G is cyclic.
To prove: \(\displaystyle\frac{{G}}{{H}}\) is cyclic.
Proof: As H is normal and G is cyclic, then G is Abelian.
Let us suppose,
\(\displaystyle{G}={\left\langle{a}\right\rangle}={\left\lbrace{a}^{{i}}{\mid}{i}\in\mathbb{Z}\right\rbrace}\)
By definition of \(\displaystyle\frac{{G}}{{H}}\), we have
\(\displaystyle\frac{{G}}{{H}}={\left\lbrace{x}{H}{\mid}{x}\in{G}\right\rbrace}\)
Use the hypothesis that G is cyclic, hence it gives
\(\displaystyle\frac{{G}}{{H}}={\left\lbrace{x}{H}{\mid}{x}\in{G}\right\rbrace}={\left\lbrace{a}^{{i}}{H}{\mid}{a}^{{i}}\in{G}\right\rbrace}={\lbrace}{a}{H}{)}^{{i}}{\mid}{a}^{{i}}\in{G}\rbrace={\left\langle{a}{H}\right\rangle}\)
Hence, it is proved that \(\displaystyle\frac{{G}}{{H}}\) is cyclic.
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