diff --git a/utils/management/__init__.py b/utils/management/__init__.py
new file mode 100644
index 0000000..e69de29
diff --git a/utils/management/commands/__init__.py b/utils/management/commands/__init__.py
new file mode 100644
index 0000000..e69de29
diff --git a/utils/management/commands/fix_proxymodel_permissions.py b/utils/management/commands/fix_proxymodel_permissions.py
new file mode 100644
index 0000000..bb880ad
--- /dev/null
+++ b/utils/management/commands/fix_proxymodel_permissions.py
@@ -0,0 +1,43 @@
+# -*- coding: utf-8 -*-
+
+"""Add permissions for proxy model.
+
+This is needed because of the bug https://code.djangoproject.com/ticket/11154
+in Django (as of 1.6, it's not fixed).
+
+When a permission is created for a proxy model, it actually creates if for it's
+base model app_label (eg: for "article" instead of "about", for the About proxy
+model).
+
+What we need, however, is that the permission be created for the proxy model
+itself, in order to have the proper entries displayed in the admin.
+
+Source: https://gist.github.com/magopian/7543724
+"""
+
+import sys
+
+from django.apps import apps
+from django.contrib.auth.management import _get_all_permissions
+from django.contrib.auth.models import Permission
+from django.contrib.contenttypes.models import ContentType
+from django.core.management.base import BaseCommand
+
+
+class Command(BaseCommand):
+    help = "Fix permissions for proxy models."
+
+    def handle(self, *args, **options):
+        for model in apps.get_models():
+            opts = model._meta
+            ctype, created = ContentType.objects.get_or_create(
+                app_label=opts.app_label,
+                model=opts.object_name.lower())
+
+            for codename, name in _get_all_permissions(opts):
+                p, created = Permission.objects.get_or_create(
+                    codename=codename,
+                    content_type=ctype,
+                    defaults={'name': name})
+                if created:
+                    sys.stdout.write('Adding permission {}\n'.format(p))