derivation-mko-first-isomorphism-theorem-finite-groups-01 · ocme-new142-reader-adoption-20261021
For f:C4 to C4,f(a)=2a mod 4, C4/{0,2} is isomorphic to {0,2}, not the full codomain C4.
Complete finite homomorphism, kernel, image, quotient and induced map with every representative checked.
calcG1 — G=Q=C4, both using the full addition-modulo-4 table below; f(0)=0,f(1)=2,f(2)=0,f(3)=2.K,I — source kernel and target image\bar f — map from cosets to the imagecalc · prerequisite — Assume exact modulo-4 arithmetic and definitions of groups, homomorphisms, quotient cosets and isomorphism, in additive notation.mko_definition · cited — Qualify both complete Cayley tables and the complete total homomorphism f first. Construct the full kernel, image and kernel-coset quotient. The induced map bar_f(g*kernel)=f(g) must agree on every representative of every coset, preserve every product and be injective and surjective onto the image. Equal cardinalities alone do not prove an isomorphism. If f is not surjective, do not replace the image by the entire codomain. Finite enumeration certifies the specified model, not the general abstract theorem; a nonhomomorphism cannot invoke the theorem. · mko-first-isomorphism-theorem-finite-groupsBefore
G=Q=C_4
After
\begin{array}{c|cccc}+&0&1&2&3\\\hline 0&0&1&2&3\\1&1&2&3&0\\2&2&3&0&1\\3&3&0&1&2\\\end{array}Conditions: none
Rule: calc · Uses: G1
Leaf basis: calc
Source and target are the same complete four-element group model, each with this 16-entry table; associativity descends from integers, identity is 0 and inverse list 0,3,2,1 is complete.
Before
f:G\to Q
After
(f(0),f(1),f(2),f(3))=(0,2,0,2)
Conditions: none
Rule: calc · Uses: G1, S01
Leaf basis: calc
All four values of the complete total map lie in Q.
Before
f((a+b)\bmod4)\equiv2(a+b)\pmod4
After
\begin{array}{c|cccc}f(a+b)&0&1&2&3\\\hline 0&0&2&0&2\\1&2&0&2&0\\2&0&2&0&2\\3&2&0&2&0\\\end{array}\quad=f(a)+f(b)\ (\bmod4)Conditions: none
Rule: calc · Uses: G1, S01, S02
Leaf basis: calc
Distributivity and reduction give equality for all 16 pairs, fully verifying the homomorphism.
Before
K=f^{-1}(0),\quad I=f(G)After
K=\{0,2\},\quad I=\{0,2\};\quad0+0=0,0+2=2,2+0=2,2+2=0\ (\bmod4)Conditions: none
Rule: calc · Uses: G1, S01, S02, S03
Leaf basis: calc
The full preimage and image are explicit; four products and inverses 0,2 qualify subgroups, and g+k-g=k for all g proves K normal.
Before
A=0+K=2+K,\quad B=1+K=3+K
After
A=\{0,2\},\quad B=\{1,3\},\quad A\sqcup B=GConditions: none
Rule: calc · Uses: G1, S01, S04
Leaf basis: calc
All four cosets deduplicate to a complete two-class partition.
Before
a,b=0,1,2,3
After
\begin{array}{c|cccc}(a+b)+K&0&1&2&3\\\hline 0&A&B&A&B\\1&B&A&B&A\\2&A&B&A&B\\3&B&A&B&A\\\end{array}\qquad\begin{array}{c|cc}+&A&B\\\hline A&A&B\\B&B&A\\\end{array}Conditions: none
Rule: calc · Uses: G1, S01, S04, S05
Leaf basis: calc
All 16 representative-pair class indices are constant on each 2-by-2 block, proving well-defined quotient addition and yielding its full four-entry table.
Before
\bar f(a+K)=f(a)
After
f(0)=f(2)=0\Rightarrow\bar f(A)=0;\quad f(1)=f(3)=2\Rightarrow\bar f(B)=2
Conditions: none
Rule: calc · Uses: G1, S02, S05, S06
Leaf basis: calc
Every representative within each coset has the same value, first proving representative independence of the induced map.
Before
A+A=A,A+B=B,B+A=B,B+B=A
After
0+0=0,0+2=2,2+0=2,2+2=0\ (\bmod4)
Conditions: none
Rule: calc · Uses: G1, S04, S06, S07
Leaf basis: calc
The images of all four quotient products match products in I, so bar f is a homomorphism.
Before
\bar f(A)=0\ne2=\bar f(B)
After
\bar f\text{ injective and onto }I=\{0,2\}Conditions: none
Rule: calc · Uses: G1, S04, S07
Leaf basis: calc
Distinct images prove injectivity, and both I elements are hit, proving surjectivity onto the image.
Before
\bar f:G/K\to I\text{ well-defined bijective homomorphism}After
G/K\cong I\subsetneq Q
Conditions: none
Rule: mko_definition · Uses: G1, S06, S07, S08, S09
Leaf basis: mko_definition
Well-definedness, operation preservation and bijectivity establish the first-isomorphism conclusion for this model.
No substeps in this guide.
S10 · premise_audit · G1, S06, S07, S08, S09 — This finite certificate does not replace the general abstract proof or substitute Q for the proper image.S09 · premise_audit · G1, S04, S07 — Codomain values 1,3 are missed, so onto Q cannot be claimed.S08 · premise_audit · G1, S04, S06, S07 — The carriers differ but all four products correspond; size two alone is not an isomorphism proof.S07 · premise_audit · G1, S02, S05, S06 — One cannot check only 0,1 and ignore 2,3; all four representatives are checked.S06 · premise_audit · G1, S01, S04, S05 — Associativity descends, identity is A, and A,B are each self-inverse, fully qualifying the quotient.S05 · premise_audit · G1, S01, S04 — Quotient elements are sets A,B rather than unique representatives 0,1.S04 · premise_audit · G1, S01, S02, S03 — K belongs to the source and I to the target, with distinct roles despite identical numeral sets.S03 · premise_audit · G1, S01, S02 — Neither cardinality nor f(0)=0 alone qualifies a homomorphism.S02 · premise_audit · G1, S01 — Q contains 1,3, which f need not hit; retain the image/codomain distinction.S01 · premise_audit · G1 — Identical group models do not make f the identity; the map still needs checking.The conclusion targets I={0,2}; without surjectivity it cannot be replaced by Q.
Check A's representatives 0,2 and B's 1,3 before the induced map.
reject — G/K has 2 elements but Q has 4, and f is not onto.reject — An actual induced map must respect all products and representatives; sizes give no operation certificate.reject — f(1)=f(3)=2, and well-definedness requires agreement.For f:C4 to C4,f(a)=2a mod 4, C4/{0,2} is isomorphic to {0,2}, not the full codomain C4.
1 declared checks for this guide; only the listed expressions are covered.
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