← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q1a-ii — Step-by-Step Solution

5 marks · Section A

Groups: definition, axioms, examples · Algebra · asked 2× in 13 yrs · Read the full method →

Question

Taking a group $\{e,a,b,c\}$ of order 4, where $e$ is the identity, construct composition tables showing that one is cyclic while the other is not.

Technique

Direct construction of both order-4 groups; identify the one containing an element of order 4 (cyclic) vs. the one where every non-identity element has order 2 (Klein).

Solution

There are (up to isomorphism) exactly two groups of order 4: the cyclic group $\mathbb Z_4$ and the Klein four-group $V_4=\mathbb Z_2\times\mathbb Z_2$.

Cyclic group $\mathbb Z_4 = \{e,a,a^2,a^3\}$

Relabel $a^2=b$, $a^3=c$. Then $a\cdot a=b$, $a\cdot b=a\cdot a^2=a^3=c$, $a\cdot c=a^4=e$, $b\cdot b=a^4=e$, etc.

Element orders: $|e|=1$, $|a|=4$, $|b|=2$, $|c|=4$.

Generators: $a$ and $c$ (both have order 4). Cyclic ✓.

Klein four-group $V_4$

Each non-identity element is its own inverse; product of any two distinct non-identity elements gives the third.

Element orders: $|e|=1$, $|a|=|b|=|c|=2$.

No element has order 4, so $V_4$ is not cyclic. ✓

Verification

• Closure, associativity (inherited from group axioms), identity ($e$), and inverses (each row/column contains $e$) hold for both tables.
• Cyclic: $a$ generates $\{a,a^2=b,a^3=c,a^4=e\}$ — all 4 elements.
• Klein: $\langle a\rangle=\{e,a\}$ only — no element generates the whole group.