title: Unable to add files to source control using VS 2015 and GIT
body: <p>One of my clients added me to their private GITHUB repo. It is a Visual Studio 2010 solution written in C#.</p>
  
  <p>I have GIT installed on my dev PC. My procedure was to load up VS 2015, go to Team Explorer -> Local Git Repositories, then Clone. I entered the GITHUB URL for this client's repo and was asked for my GITHUB username/password.</p>
  
  <p>Everything seemed to work. I see no errors. The progress bar displayed as expected and the local repo was created. Now I can see the solution and all of its projects. I am able to build the solution and even run/debug it.</p>
  
  <p>I see padlocks to the left of files. If I edit one of these files I will see a red check mark indicating the file is checked out.</p>
  
  <p>The problem I am experiencing is that I can't add a file to the solution and include it in source control. Normally I can drag/drop or copy/paste files and they are added to source control. I have even tried adding item by right clicking. No matter what I do, the file is added but there is no icon to the left of it to indicate its source control status - of course, I am expecting a <code>+</code> sign. Of course, the added file is also not showing up under Changes.</p>
  
  <p>Another symptom is if I rename an existing file. At first it has a padlock. When I rename the file it will not have a status icon to it at all.</p>
  
  <p>I've read a ton of posts from people with a similar problem, but none of those solutions are effecting mine. I'm not seeing any errors, so I don't know what would be relevant to include in this post. :S</p>
  
  <p><strong>EDIT</strong></p>
  
  <p>I am obviously not a GIT guru... but I do understand that the padlock is telling me <code>read only</code> (I think). I don't know why and haven't been able to find a clue. Thought someone might point me in a good direction. When I add a file I do see the <code>+</code> sign very momentarily, but then it disappears and no errors are shown in output window.</p>