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Always true. The synthetic definition identifies both sides with the order type obtained by concatenating well-orders of types $\alpha$, $\beta$, and $\gamma$ in that order. Thus ordinal addition is associative:

The inductive definition of ordinal addition is the transfinite recursionfor every nonzero limit ordinal $\lambda$. The synthetic definition takes disjoint well-orders $A$ and $B$ of order types $\alpha$ and $\beta$, puts every member of $A$ before every member of $B$, and defines $\alpha +\beta$ to be the order type of this ordered sum.

These definitions agree by transfinite induction on $\beta$. The empty second order contributes nothing; adjoining a greatest element gives the successor clause; and a limit well-order is the union of all of its proper initial segments, so its ordered sum with $A$ has order type ${\sup }_{\gamma <\lambda }(\alpha +\gamma )$.

The inductive definition of ordinal addition isfor a limit ordinal $\lambda$.

The synthetic definition takes the order type of the disjoint union of a copy of $\alpha$ followed by a copy of $\beta$: every point of the first copy precedes every point of the second, and each copy retains its original total order.

Fix $\alpha$ and use transfinite induction on $\beta$. The synthetic sum with the empty order is $\alpha$. Appending a greatest element produces the successor ordinal rule. At a limit ordinal $\lambda$, the second copy is the union of its initial segments of types $\beta <\lambda$, so the whole order has type
${\sup }_{\beta <\lambda }(\alpha +\beta )$. Thus the synthetic operation satisfies the inductive recursion; uniqueness in transfinite recursion proves the definitions equivalent.

True. The hypothesis says that $\alpha +\alpha$ and $\beta$ commute under ordinal addition. By the commuting ordinal addition classification, comparison of their Cantor normal forms implies that $\alpha$ and $\beta$ themselves commute. Thereforeso both outer expressions equal the middle one.

First let $\delta ={\omega }^{\alpha }$. If $\beta ,\gamma <\delta$, every exponent appearing in their Cantor normal forms is strictly below $\alpha$. The rules for ordinal addition can delete lower terms but cannot introduce an exponent at least $\alpha$. Thereforeso $\delta$ is additively closed.

Conversely, suppose nonzero $\delta$ is additively closed. Part (a) givesIf $\rho >0$, then both ${\omega }^{\alpha }n$ and $\rho$ are smaller than $\delta$, but their sum is $\delta$, contradicting closure. Hence $\rho =0$. If now $n>1$, thenwhileanother contradiction. Thus $n=1$, andThis proves the classification of additively closed ordinals.

Suppose first that $\delta >2$ is multiplicatively closed. It is also additively closed. Indeed, for $\beta ,\gamma <\delta$, let $\mu =\max \{\beta ,\gamma \}$. Monotonicity of ordinal addition giveswhere the last inequality uses $\mu <\delta$, $2<\delta$, and multiplicative closure. Part (b) therefore givesfor some nonzero ordinal $\lambda$.

For any $\rho ,\sigma <\lambda$, strict monotonicity givesMultiplicative closure and the exponent law now implyhence $\rho +\sigma <\lambda$. Thus $\lambda$ is additively closed, and part (b) gives $\lambda ={\omega }^{\alpha }$. Consequently

Conversely, let $\delta ={\omega }^{\lambda }$ with $\lambda ={\omega }^{\alpha }$. By part (b), $\lambda$ is additively closed. Take nonzero $\beta ,\gamma <\delta$, with leading exponents $\rho ,\sigma <\lambda$ in Cantor normal form. If $\gamma$ is finite, the product $\beta \gamma$ has leading exponent $\rho <\lambda$. If $\gamma$ is infinite, the leading exponent of an ordinal product isby additive closure of $\lambda$. In either caseProducts involving zero are immediate, so $\delta$ is multiplicatively closed. This is the multiplicative closure criterion for a power of omega and completes both directions.