11 Groups

Order of an element¶

For \(+, \oplus\)¶

Regular/modulo addition operator

\(x \oplus y =\) ??

what positive number multiplied gives the product as the identity element

For \(\times, \otimes\)¶

Regular/modulo multiplication operator

what number raised to gives the final answer as the identity element

For Both¶

If \(a \in G\), G is a group, then the order of \(a\) is the order of the cyclic group

COSETS & Lagrange’s Theorem¶

Let

• \(G\) be a group
• \(H\) be its subgroup
• \(a \in G\)

• \(h \in H\)

• COSETS may be duplicated, but we are only concerned about disjoint COSETS

• Union of disjoint COSETS will be \(G\)
• no of elements in COSET = no of elements in \(H\)

Theorems¶

If \(b \in G, b \ne a\)

1. \(a \in H \iff aH = H\)
2. \(aH = bH \iff a^{-1} b \in H\)
3. \(a \in bH \iff a^{-1} \in H b^{-1}\)
4. \(a \in bH \iff aH = bH\)

\(^{-1}\) means inverse (could be additive or multiplicative inverse)

Lagrange’s Theorem¶

Let \(G\) be a finite group of order \(n\) and \(H\) be any subgroup of \(G\). Then the order of H divides the order of \(G\).

\([G:H]\) = index of H = no of distinct left COSETS of H in G

Let \(r\) be index of H. Let \(|G| = n,|H| = m\). Then \(n = mr \implies \frac n m = r\). Clearly, \(m\) divides \(n\)

Application¶

A cyclic group can only have subgroups with no of elements which divides the

eg: \((Z_7, \oplus_7)\) can only have subgroups having no of elements dividing \(7\). So, it can either be \(<1>\) or \(<7>\).

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