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An automatic group admits an automatic structure for a group: a regular language of representatives and synchronous group multiplier automata recognizing right multiplication by each generator and equality of representatives. Representatives need not be unique. This finite-state description yields effective algorithms for the word problem for a group.

An automatic structure for a group consists of a finite generating alphabet $A$, a surjective representative regular language $L\subseteq A^{\ast }$, and finite-state automata accepting the padded convolutions of words $(u,v)$ in $L$ satisfying $\overline{u}x=\overline{v}$ for each $x\in A\cup \{\epsilon \}$. Symmetry of $A$ and uniqueness of representatives are additional conditions, not part of this definition.

Use the synchronous group multiplier automata belonging to the automatic structure for a group. Write $u\otimes v$ for the padded convolution of words: the shorter tape is completed with a symbol $\#$ outside the alphabet, and reading stops when both words over an alphabet have ended. Our convention isLet $N\ge 1$ be at least the number of states of each of these finite-state automata, including $M_{\epsilon }$.

Choose a shortest representative $v\in L$ of $g$. In the first orientation, $M_x$ accepts $w\otimes v$; in the other orientation it accepts $v\otimes w$. If $|v|>|w|+N$, the accepting run has more than $N$ transitions after the tape containing $w$ has ended. Two states in that suffix coincide. Delete the intervening nonempty loop. Every deleted column contains $\#$ on the fixed tape and a genuine letter on the tape containing $v$, so the resulting input remains a valid padded convolution of words. The group multiplier automaton still accepts, giving a shorter representative of the same $g$. This contradicts minimality. Thus a representative can always be chosen withThis argument does not require a symmetric alphabet: the second orientation uses the same $M_x$ with its tapes interchanged in the input, rather than a presumed $M_{x^{-1}}$.

Fix the first tape of the group multiplier automaton $M_x$ to be $u=a_1\cdots a_m$. Construct its possible runs by dynamic programming, one column at a time. A record consists of a state and a flag indicating whether the output tape has ended. At column $i\le m$, the first symbol is $a_i$ and the second symbol may be any letter of the alphabet, or $\#$; after the second tape first uses $\#$ it must continue to use $\#$. At later columns the first symbol is $\#$, and an additional column must have a genuine letter on the second tape. Never add a $(\#,\#)$ column.

Keep one predecessor record and its output symbol for each reachable state-and-flag pair in each layer. Two runs reaching the same record have exactly the same possible future completions, so this merging loses no accepting output. The merged dead state of a finite automaton can be discarded. At each layer $i\ge m$, test whether a reached state is accepting; the initial layer is included when $m=0$. Upon success, follow predecessor records backwards and omit padding symbols to recover $v$.

The automatic structure for a group supplies a representative in $L$ of $\overline{u}x$, so an accepted padded convolution of words exists and this search terminates. The output is the required representative:The search uses only the finite transition table; it does not presuppose a solution to the word problem for a group.

Part (a)(i), applied to $w=u$ and $g=\overline{u}x$, guarantees an accepted output of length at most $m+N$. Its padded convolution of words has at most $m+N$ columns. Thus the dynamic programming search in part (b) reaches an accepting layer by this bound.

Each layer has at most twice the fixed number of states of $M_x$, and each record has at most $|A|+1$ outgoing choices. These are constants for the fixed automatic structure for a group. Storing a pointer and a symbol takes constant time; copying whole partial output strings at every step would unnecessarily spoil this bound. Reconstruction also takes at most $m+N$ steps. Therefore automatic multiplication has linear time and bounded length increase:For nonempty inputs this is the requested $O(|u|)$ time complexity. The additive $1$ merely includes the constant work on the empty word.

The regular language $L$ is nonempty because it represents every element of the group. Find any accepted word $t=a_1\cdots a_k$ using Breadth-first search in its finite-state automaton. If $t$ is the empty word, it already represents the identity.

Otherwise begin with $t\in L$ and repeatedly apply the linear-time multiplication in an automatic structure to the lettersThese letters belong to $A$ by symmetry of the alphabet. All intermediate words lie in $L$, and the final word $\gamma$ representsHence an identity representative can be constructed effectively:This is a finite preliminary computation depending only on the fixed automatic structure for a group; it uses no test for whether an arbitrary word represents the identity.

Write the input word over an alphabet as $w=a_1\cdots a_n$. Starting with $z_0=\gamma$, obtain $z_i$ from $z_{i-1}$ by linear-time multiplication in an automatic structure with $a_i$. Mathematical induction givesThe time complexity of step $i$ is at most a constant times $|\gamma |+(i-1)N+1$. Summing the costs givesThe automatic structure for a group, $N$ and $\gamma$ are fixed, so the required representative isFor the empty word, return $\gamma$ in constant time.

Construct $z$ as above, and run the equality group multiplier automaton $M_{\epsilon }$ on $z\otimes \gamma$. Both inputs belong to $L$, soThe final finite-state automaton run takes $O(\max (|z|,|\gamma |)+1)=O(n+1)$ time. The word problem for a group is therefore decidable in quadratic time:The empty word is accepted immediately. Comparing $z$ and $\gamma$ as literal strings would be incorrect, since an automatic structure for a group need not give unique representatives.