Few doubts about direct products of special groups

Question

Let $G_1$ be a direct product of three groups $A$,$B$ and $Q$ such that $A$ is an abelian group whose order is odd (let's say $k$ ) and $B$ is an abelian group whose order is $2^m$ and which is direct product of 2-groups, $C$ is quaternion group of order 8. let $H$ be a subgroup of $G_1$, now take $h =abc$, where $a\in A$, $b\in B$ and $c \in Q$. let $g$ be any element of $G_1$ and write $g = \alpha \beta\gamma$, where where $\alpha\in A$, $\beta \in B$ and $\gamma\in Q$. Then we have $g^{-1}hg = \gamma^{-1}\beta^{-1} \alpha^{-1}abc\alpha\beta\gamma = \gamma^{-1}abc\gamma = \gamma^{-1}c\gamma ab$. since $c^4 = 1$ and $c$ is self conjugate in $Q$, it follows that $\gamma^{-1}c\gamma = c$ or $c^3$(why?). Hence $g^{-1}hg = h$ or $c^2h$. Therefore $H$ contains both $h$ and $c^2h$ and hence contains $g^{-1}hg$. Therefore $H$ is self-conjugate in $G_1$.

Question1 : $\gamma^{-1}c\gamma = c$ or $c^3$(why?)

Question 2 : $g^{-1}hg = \gamma^{-1}\beta^{-1} \alpha^{-1}abc\alpha\beta\gamma = \gamma^{-1}abc\gamma = \gamma^{-1}c\gamma ab$ (I am not getting how?)

Question 3: Why $c^4 =1$ (why they are considering only order 4 elements of $Q$ I mean there is an element of order 2 also in it)?

Answer 1

Questions 1 and 3 are essentially questions about the quaternion group.

Every element $c$ of the quaternion group has order dividing $4$, so $c^4=1$. (Keep in mind that, for a group element $x$, $x^n=1$ does not mean $x$ has order $n$, it only means $x$ has order dividing $n$.)

Similarly, the fact that $\gamma^{-1} c\gamma\in \{c,c^3\}$ is true for every pair of element $\gamma,c\in Q$. You can simply check: if $c$ does not have order $4$, then it is central and has order dividing $2$, so $c^3=c$ and we're fine. If $c$ has order $4$, one can check that its conjugacy class is exactly $\{c,c^3\}$.

As for Question 2, note that $a$ commutes with each of $\alpha,\beta,\gamma$. (Here you use the fact that some of these groups are abelian, and properties of direct products). So does $b$. $c$ also commutes with $\alpha$ and $\beta$, but not necessarily with $\gamma$. With a little effort, you should be able to see why the equality holds from this.

Answer 2

In the quaternion group $\pm 1$ are central. Their only conjugates are themselves. The other elements are like $i$, whose only conjugates are $\pm i$ and $-i=i^3$.

For question 2, $A$ and $B$ are Abelian. Also as the product is direct, all elements in $A$ and $B$ commute with all elements of $C$.

All elements of the quaternion group satisfy $c^4=1$, even those whose orders are less than $4$.