Let
$$G=\left\{\begin{pmatrix}a & b\\0& 1\end{pmatrix}:a>0,b\in \mathbb{R}\right\}$$
be endowed with matrix multiplication and the inherited topology as a subspace of $\mathbb{R}^4$. Show that $G$ is a locally compact group.
A locally compact group is a topological group whose topology is locally compact and Hausdorff. So, I'd have to show that. However, I would also have to show that $G$ is a topological group. $G$ is a group, which is easy to show; but I would need to show that $(x,y)\rightarrow xy$ and $x\rightarrow x^{-1}$ are continuous (in this case, it should be, I believe, $(w,x,y,z)$ since we are in a subspace of $\mathbb{R}^4$). Long story short, I am confused which direction to go. I'd suspect that there should be a relatively straightforward easy way to show that $G$ is a locally compact group, because from what I've seen, that is typically just assumed. I found this, but unfortunately didn't gain much insight from it.
I'd appreciate any guidance/hint. Thanks.
You're right that showing that this group is locally compact isn't too bad. I'll assume that you're OK with the continuity of the group operations, these are just rational/linear functions so it should be clear that they are continuous.
To show that the topology is Hausdorff is inherited from the topology of $\mathbb{R}^4$. If $A,B \in \mathbb{R}^4$ then there exist open sets $A \in U,B \in V$ with $U \cap V = \emptyset$. Now instead of arbitrary elements of $\mathbb{R}^4$ assume that $A,B \in G$. The sets $U\cap G,V \cap G$ are open sets in the topology of $G$ that appropriately separate $A$ and $B$ so $G$ is Hausdorff.
Local compactness is also inherited but here is a long-winded argument. First consider the subset of $\mathbb{R}^4$, $S = \{(x,y,1,0):x,y, \in \mathbb{R}^2\}$, which is closed. This subset of $\mathbb{R}^4$ will be locally compact in its subspace topology in light of the fact that intersections of compact sets with closed sets are compact (see **). Now look at the subspace of $S$, $G$. Notice that $G$ is an open set in $S$. Let $C$ be a compact neighbourhood in $S$ of a point $A \in G$. Thus we can find an open ball $B_{\delta}(A) \cap S$ in $S$ about $A$ that is contained in $G$ and whose closure is contained in $C$. Then $cl(B_{\delta(A)} \cap S) \subset G$ is a compact set in $S$ (being closed and contained in the compact set $C$ -- note we need that $S$ is Hausdorff for this). In addition, $cl(B_{\delta(A)} \cap S)$ contains the set $B_{\delta(A)} \cap S$ which is open in $G$. To see that $cl(B_{\delta(A)} \cap S)$ is actually compact in the topology of $G$ you can revert to the open cover definition of compactness and cover it with open sets of $G$ and see if you can find a finite subcover. The compactness of $cl(B_{\delta(A)} \cap S)$ in $S$ will help you here (see ** again).
(**) Basically all you need to show is that if a set, $C$ is compact in a topological space ,$X$, and $C \subset Y \subset X$ then $C$ is compact in the subspace topology of $Y$.
Rather than go through the above mess of an argument it is easier to show local compactness constructively: for $(a,b,1,0) \in G$ the set $C = \{(x,y,1,0): x \in [\frac{a}{2},\frac{3a}{2}], y \in [y-1,y+1]\}$ is a compact neighbourhood of $G$.
In general showing something like local compactness for a matrix group shouldn't give you too much grief. Usually the group is either an open of closed set of $\mathbb{R}^n$. In this case this matrix group should be viewed as an open set of $\mathbb{R}^2$.