I was studying about group theory when I came across a very well-known statement that:
Any two dihedral groups of degree $4$ are isomorphic.
I have two questions from here:
What’s actually is a degree of a group? I have heard about the order of a group, but never heard anything about degree of a group.
Secondly, how are they concluding the fact that any two dihedral groups of degree $4$ are isomorphic?
I know that a group is called a “dihedral group of degree $4$” if it has only two generators $a, b$ such that $o(a) = 4$, $o(b) = 2$ and $ba = a^3b$. From these facts I proved that a dihedral group of degree $4$ always has eight elements.
My proof for my above assertion goes as follows:
If $a, b$ are two generators of the above group then we have $G = \langle a, b \rangle$. Hence we can say $$ G = \{ a^{i_1} b^{j_1} a^{i_2} b^{j_2} \dotsm a^{i_n} b^{j_n} \mid i_t, j_t \in \mathbb{Z}, 1 \leq t \leq n, n \in \mathbb{N} \} $$ Now, we have, $b a = a^3 b$ due to which every element of $G$ is of the form $a^i b^j$, where $i, j \in \mathbb{Z}$. Since $a^4 = e$ and $b^2 = e$, so $G = \{ a^i b^j \mid 0 \leq i \leq 4, 0 \leq j \leq 2\}$. Now, $|G| \leq 8$.
Since, $o(a) = 4$, so $P = \{ e, a, a^2, a^3 \}$ are distinct. Also, since $a \neq b \neq e$, hence $Q = \{ b, a b, a^2 b, a^3 b\}$, are distinct as well. Now we claim $P \cap Q = \emptyset$. This is because $b \neq e$, $b \neq a$, if $b = a^2$, then $a^3 b = b a b = b a$, due to which $b = e$, a contradiction. Also, if $b = a^3$, then $a^3 b = b^2 = b a$, then $a = b$, a contradiction again. Now, if $ab = e$, then $a^3b = a^2 = ba$, a contradiction. Also, $a b \neq a$, $a b \neq a^2$. If $a b = a^3$, then $a^2 = b$, a contradiction shown in a previous case. Also, from previous cases we can say $a^2b \neq e, a, a^2, a^3$. Similarly, $a^3b \neq e, a, a^2, a^3$. Hence $G = P \cup Q = \{e, a, a^2, a^3, b, a b, a^2 b, a^3b \}$. Hence, $|G| \leq 8$ and $|G| = 8$.
This completes the proof. But I don’t know how to show that:
Any two dihedral groups of degree $4$ are isomorphic.
I think that if we consider two dihedral groups of degree $4$ say , $D_4$ and $D'_4$ such that $D_4=\{e, a, a^2, a^3, b, a b, a^2 b, a^3b \}$ and $D'_4=\{\overline e, p, p^2, p^3, r, p r, p^2 r, p^3r \}$, and consider the following bijective mapping $f:D_4\longrightarrow D_4'$ such that $f(e)=\overline e,f(a)=p,f(a^2)=p^2,\cdots ,f(a^3b)=p^3r$, then we just need to show that $f$ is a homomorphism . Now, I dknt get how to show this as well . One way is there if we form each pairs of $(a^ib^j,a^cb^d)$ and compute , but that becomes an unusually long process as we have to do $64$ computations manually ...
One way to do this is to notice that the information you have completely determines the multiplication table for the group.
Indeed, you know that the group is generated by $a$ and $b$ and has $o(a)=4$, $o(b)=2$ and $ba=a^3b$.
As you observed, from the last equality alone we can conclude that $G=\{a^ib^j:i,j\in\mathbb Z\}$, and using the fact that $a$ and $b$ have orders $4$ and $2$, respectively, we can improve this description to $$G=\{a^ib^j:0\leq i<4,0\leq j<2\}.$$ Moreover, then $8$ elements listed here are pairwise different. Indeed, let us suppose that $i$, $i'$, $j$ and $j'$ are integers such that $0\leq i,i'<4$, $0\leq j,j'<2$ and $a^ib^j=a^{i'}b^{j'}$. If $j\neq j'$, then we can suppose without loss of generality that $j=0$ and $j'=1$, so the equality $a^ib^j=a^{i'}b^{j'}$ tells us that $a^{i-i'}=b$: now $a^3b=ba=a^{i-i'}a=aa^{i-i'}=ab$, so $a=a^3$, and this is absurd, since $a$ has order $4$. It follows that $j=j'$ and, since $a^ib^j=a^{i'}b^{j'}$, that $a^i=a^{i'}$: as the order of $a$ is $4$, this implies that $i-i'$ is divisible by $4$, and since both $i$ and $i'$ are in $\{0,1,2,3\}$, that they are actually equal.
The conclusion of all this is that $G$ has exactly $8$ elements, which are $$a^ib^j,\qquad\text{with $0\leq i<4$ and $0\leq j<2$.}$$ Now using the equality $ba=a^3b$ we can easily check that $ba^i=a^{3i}b$ for all $i\in\mathbb Z$, and using that, in turn, that $$a^ib^j\cdot a^{i'}b^{j'}=a^{i+3i'}b^{j+j'}$$ for all choices of $i$, $i'$, $j$ and $j'$ in $\mathbb Z$. This tells us how to write down the complete multiplication table of $G$.